국제바칼로레이트 · IB Diploma · Physics · HL

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A speedometer gives a reading at one moment. A journey average can hide stops and rapid changes in speed.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Displacement includes direction; distance counts total path length. Velocity 速度 is change in displacement per time. Acceleration is change in velocity per time.
• velocity: Rate of change of displacement; acceleration: Rate of change of velocity.

Choose evidence that can test it

• The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement. A constant-acceleration formula is valid only when its assumption is justified.
• Choose a positive direction and state it. Use a light gate or video with a known scale and frame interval for repeatable motion measurements. Avoid assuming hand timing is exact over very short intervals.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: velocity rises uniformly from 2 to 10 metres per second in 4 s. a = (v-u)/t. a = (10-2)/4 = 2 metres per second squared. Displacement is trapezium area: s = (u+v)t/2 = (2+10)×4/2 = 24 m.

Example:

Velocity changes from 3 to 15 metres per second in 6 s. Find acceleration. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A negative velocity always means the object is slowing down. This claim is false: Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.

For uniformly accelerated motion 匀变速运动 the guide gives four equations — derive or verify each from a velocity–time graph, then choose by what you know:

Worked check. A stone is dropped from a 45 m cliff (g = 9.8 m/s², down positive): time from $s = \tfrac12 gt^2$: $t = \sqrt{2 \times 45/9.8} = 3.03$ s; landing speed $v = gt = 9.8 \times 3.03 = 29.7$ m/s.

With no fluid resistance, resolve into independent horizontal (constant velocity) and vertical (constant acceleration) motions; the trajectory is parabolic (its equation is not required).

Worked check. A dart is thrown horizontally at 8.0 m/s from 1.8 m above the floor: fall time $t = \sqrt{2 \times 1.8/9.8} = 0.606$ s; horizontal distance $= 8.0 \times 0.606 = 4.85$ m; vertical component at impact $= 9.8 \times 0.606 = 5.94$ m/s, so speed $= \sqrt{8.0^2 + 5.94^2} = 9.96$ m/s.

With fluid resistance, the path is no longer parabolic and the range is reduced; the change in flight time depends on the launch conditions, and a falling object approaches its terminal speed 终端速度 when drag balances weight — qualitative effects only.

Key:

Motion graphs and acceleration: The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement. A constant-acceleration formula is valid only when its assumption is justified.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A passenger continues moving when a vehicle brakes. The seat belt provides the force needed to change the passenger momentum.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Resultant force 合力 causes acceleration, not motion itself. Momentum is mass multiplied by velocity. For an isolated system, total momentum is conserved even when kinetic energy is not.
• momentum: Mass multiplied by velocity; resultant force: The vector sum of forces on an object.

Choose evidence that can test it

• Impulse equals momentum change. Increasing stopping time for the same momentum change reduces average force. Identify external forces before applying momentum conservation.
• Draw a free-body diagram containing only forces on the selected object. For spring measurements, add loads in steps within the elastic range and measure extension from the unloaded position.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 60 kg passenger slows from 10 to 0 metres per second in 0.50 s. Δp = m(v-u) = 60(0-10) = -600 kg metres per second. Average F = Δp/Δt = -600/0.50 = -1,200 N. The sign shows the force opposes the initial motion.

Example:

A 2 kg trolley travels at 3 metres per second. Find its momentum. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Balanced forces do not require the object to be at rest. Mass and weight have different units and meanings.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

An object moving at constant velocity must have a nonzero resultant force. This claim is false: Balanced forces do not require the object to be at rest. Mass and weight have different units and meanings.

The guide names each force with a symbol and, where it applies, an equation. Contact forces 接触力 act where bodies touch: the normal force $F_N$ acts perpendicular to the surface; friction 摩擦力 $F_f$ acts parallel to the contact plane — static $F_f \le \mu_s F_N$ (up to a limit) and dynamic $F_f = \mu_d F_N$; tension 张力 pulls along a rope or string; an elastic restoring force follows Hooke's law $F_H = -kx$; viscous drag on a small sphere is $F_d = 6\pi\eta rv$; and buoyancy 浮力 is $F_b = \rho V g$ on the displaced fluid's volume. Field forces 场力 act without contact: weight $F_g = mg$, electric force $F_e$, magnetic force $F_m$.

Worked check (friction). A 2.0 kg block slides on a table ($\mu_d = 0.30$, g = 9.8 m/s²): $F_f = 0.30 \times 2.0 \times 9.8 = 5.9$ N. A 5 N horizontal pull cannot even start it if $\mu_s = 0.40$: the static limit is $0.40 \times 19.6 = 7.8$ N.

Free-body diagrams isolate one body and show every force on it; resolve to find the resultant. Newton's first law: zero resultant means constant velocity (translational equilibrium 平移平衡). Second law: $F = ma$ for constant mass, or more generally:

The impulse 冲量 $J$ delivered by a resultant force equals the momentum change it causes. Total momentum stays constant unless a resultant external force acts — so collisions and explosions conserve momentum in isolated systems.

Worked check. A 0.15 kg ball arrives at 12 m/s and leaves at 12 m/s the opposite way: $\Delta p = 0.15 \times 24 = 3.6$ kg·m/s; if contact lasts 0.020 s the average force is $3.6/0.020 = 180$ N. The guide treats quantitative collisions in one dimension at SL and two dimensions at HL; simultaneous energy-and-momentum equation pairs are not required.

Key:

Forces, momentum and safe stopping: Impulse equals momentum change. Increasing stopping time for the same momentum change reduces average force. Identify external forces before applying momentum conservation.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A motor can transfer some input energy to lifting and the rest to heating. Useful output is part of the total energy transfer.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Work done by a constant force parallel to displacement is force multiplied by distance. Kinetic energy depends on speed squared. Energy is conserved when all transfers and stores are included.
• power 功率: Energy transferred per unit time; efficiency: Useful output divided by total input.

Choose evidence that can test it

• Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.
• Measure a lifting height and load, time the lift, and record electrical input with suitable instruments. Repeat trials and account for heating or friction as transfers, not missing energy.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a device receives 600 J and transfers 420 J usefully. Efficiency = useful output / total input. Efficiency = 420/600 = 0.70 = 70%. Over 3 s, useful power = useful energy/time = 420/3 = 140 W.

Example:

A device receives 250 J and gives 150 J useful output. Find efficiency as a percentage. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Dissipated energy is destroyed. This claim is false: Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.

Work done by a constant force depends on the component of the force along the displacement:

The net work done on a particle equals its change in kinetic energy. Mechanical energy is kinetic + gravitational potential + elastic potential; without frictional or resistive forces it is conserved, and work is the amount transformed between mechanical forms:

Worked check. A 2.0 kg crate is pulled 5.0 m by a 20 N rope at 30° above the horizontal: $W = 20 \times 5.0 \times \cos 30° = 86.6$ J. If it starts at rest on a frictionless floor, $E_k$ gained is 86.6 J, so $v = \sqrt{2 \times 86.6/2.0} = 9.3$ m/s.

Power is the rate of energy transfer, and efficiency compares output with input:

Worked check (power two ways). A car drives at a steady 25 m/s against 600 N of resistive force: $P = Fv = 600 \times 25 = 1.5 \times 10^4$ W. If the engine inputs 20 kW, $\eta = 15/20 = 0.75 = 75\%$.

The guide adds energy density of fuel sources as a comparing quantity (energy per unit mass or volume of fuel). The guidance point to carry into evaluation: interpret any change in total mechanical energy as the work done on the system by non-conservative forces — friction, drag, driving forces that are not stored-and-returnable.

Key:

Energy stores, work and efficiency: Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.

Full A.4 Rigid body mechanics coverage (HL ONLY — the guide states there is no standard-level content in A.4), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 36–37). Vector treatment of torque/angular momentum is not required; senses (clockwise/counter-clockwise) are. Rolling without slipping only for combined motion.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A door is easier to open when pushed near its outer edge. Force size alone does not determine the turning effect.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Torque 力矩 depends on force and perpendicular distance from the axis. Moment of inertia 转动惯量 describes resistance to angular acceleration and depends on mass distribution about a stated axis.
• torque: Turning effect about an axis; moment of inertia: Resistance to angular acceleration about a specified axis.

Choose evidence that can test it

• For a rigid body about a fixed axis, net torque equals moment of inertia multiplied by angular acceleration. Rotational kinetic energy depends on angular speed squared. Angular momentum is conserved when net external torque is zero.
• Draw the pivot, force direction and perpendicular lever arm. Do not use the sloping distance from pivot to force point unless it is perpendicular to the force. For experiments, keep rotating parts guarded and loads secure.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 10 N perpendicular force acts 0.30 m from an axis. Torque = force × perpendicular distance = 10×0.30 = 3.0 N m. If moment of inertia is 0.60 kg square metres, angular acceleration = torque/inertia = 3.0/0.60 = 5.0 radians per second squared.

Example:

A 12 N force acts perpendicularly 0.25 m from the pivot. Find torque. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Torque has unit N m but is a turning effect, not an energy store. Moment of inertia changes if the same mass moves farther from the axis.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Moment of inertia depends only on total mass, never its distribution. This claim is false: Torque has unit N m but is a turning effect, not an energy store. Moment of inertia changes if the same mass moves farther from the axis.

Torque is the rotational analogue of force; with the sense (clockwise or counter-clockwise) stated:

The moment of inertia depends on how mass is distributed about the axis; for point masses $I = \sum mr^2$. The angular equations of motion mirror the linear ones:

Angular momentum and rotational energy complete the analogy (constant unless a resultant torque acts; rolling without slipping couples $v = \omega r$):

Worked check (see-saw). A 300 N child sits 1.2 m left of the pivot: torque $= 300 \times 1.2 = 360$ N·m counter-clockwise; a 400 N child 0.90 m right gives $360$ N·m clockwise — equilibrium.

Worked check (point-mass inertia). Three 0.50 kg masses at the corners of an equilateral triangle of side 0.60 m, each $r = 0.60/\sqrt3 = 0.346$ m from the centre: $I = 3 \times 0.50 \times 0.346^2 = 0.18$ kg·m². A 0.20 N·m torque gives $\alpha = \tau/I = 0.20/0.18 = 1.1\ \mathrm{rad/s^2}$.

Worked check (changing I). A skater with $I = 4.5$ kg·m² spins at $\omega = 2.0$ rad/s; pulling her arms in halves $I$ to 2.25. No external torque, so $L = 9.0$ kg·m²/s is conserved: $\omega' = 9.0/2.25 = 4.0$ rad/s, and the rotational energy rises from $\frac12 \times 4.5 \times 4 = 9.0$ J to $\frac12 \times 2.25 \times 16 = 18$ J — her muscles did the extra work.

Key:

Rigid bodies and rotational dynamics: For a rigid body about a fixed axis, net torque equals moment of inertia multiplied by angular acceleration. Rotational kinetic energy depends on angular speed squared. Angular momentum is conserved when net external torque is zero.

Full A.5 Galilean and special relativity coverage (HL ONLY), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 37–38). Derivations of Lorentz/velocity-addition/dilation equations are not required; their use is. World lines limited to constant velocity; time axis labelled ct.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Two observers can assign different times between the same events. A clock reading has meaning only when the frame and events are specified.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• In special relativity, inertial observers measure the same vacuum light speed. Time dilation compares proper time 固有时间 measured where two events occur at one place with the interval in another inertial frame 惯性参考系.
• proper time: Time between events measured at one location in a frame; inertial frame: A frame in which a free body moves at constant velocity.

Choose evidence that can test it

• The Lorentz factor is 1 divided by the square root of 1 minus speed squared over light speed squared. Galilean velocity addition is an approximation for speeds much smaller than light speed.
• Label which frame measures proper time or proper length before substituting. Use event coordinates consistently. Do not combine lengths from one frame with time intervals from another without transformation.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: v = 0.60c and proper time is 4.0 microseconds. γ = 1/√(1-v²/c²) = 1/√(1-0.60²) = 1.25. Δt = γΔt₀ = 1.25×4.0 = 5.0 microseconds in the stated second frame.

Example:

A proper interval is 8 microseconds and gamma is 1.25. Find the other-frame interval. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Time dilation is not an instrument fault. Proper length is measured in the object rest frame; a contracted length is measured in a frame where it moves.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

All inertial observers must measure the same time interval between any two events. This claim is false: Time dilation is not an instrument fault. Proper length is measured in the object rest frame; a contracted length is measured in a frame where it moves.

A reference frame is inertial when it does not accelerate. Newton's laws hold in all inertial frames (Galilean relativity), giving the everyday transformations $x' = x - vt$, $t' = t$ and the velocity addition $u' = u - v$. Special relativity — from its two postulates (the laws of physics are the same in all inertial frames; light's speed is the same in all of them) — replaces these with the Lorentz transformations:

and the relativistic velocity addition $u' = (u - v)/(1 - uv/c^2)$. The space–time interval is invariant: $(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$.

Worked check (gamma). At $v = 0.60c$: $\gamma = 1/\sqrt{1 - 0.36} = 1/0.80 = 1.25$.

Proper time $\Delta t_0$ is measured by a clock at both events; proper length $L_0$ by a ruler at rest in the object's frame:

Worked check (muon). A muon with proper lifetime 2.2 µs travels at 0.995c ($\gamma \approx 10.0$): lab lifetime $= 10.0 \times 2.2 = 22$ µs, and in the muon's frame Earth's atmosphere is contracted by the same factor — both views agree it reaches the ground. Muon decay experiments are the guide's named evidence.

On a space–time diagram (time axis labelled $ct$), a particle's world line at constant velocity makes an angle with the time axis: $\tan\theta = v/c$; simultaneity itself is frame-dependent. Derivations of the transformation, dilation and contraction equations are not required — their use is.

Key:

Relativity and measuring events: The Lorentz factor is 1 divided by the square root of 1 minus speed squared over light speed squared. Galilean velocity addition is an approximation for speeds much smaller than light speed.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Two equal masses receive the same energy but show different temperature rises. Material properties determine how energy transfer changes temperature.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Specific heat capacity 比热容 is energy needed to raise the temperature of unit mass by one degree. Specific latent heat 潜热 relates energy to change of state without temperature change for the idealized process.
• specific heat capacity: Energy per mass per temperature rise; latent heat: Energy associated with a change of state.

Choose evidence that can test it

• Temperature relates to particle motion in a model; internal energy includes kinetic and potential contributions. During a change of state, energy can change particle arrangements rather than temperature.
• Measure mass, electrical input and temperature change for an insulated block. Ensure the temperature sensor has good contact, allow time for equilibration, and consider energy transferred to the surroundings.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 0.50 kg block gains 2,000 J and rises 10 °C. Use E = mcΔT. Rearranging gives c = E/(mΔT). c = 2,000/(0.50×10) = 400 J per kilogram per degree. Heat loss would make the value inferred from electrical input too large.

Example:

A 2 kg sample with c=500 J per kilogram per degree warms by 3 °C. Find energy gain. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A flat section of a heating curve can show a phase change, not absence of energy transfer. Do not substitute temperature for a temperature difference in E = mcΔT.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A temperature plateau always means zero energy transfer. This claim is false: A flat section of a heating curve can show a phase change, not absence of energy transfer. Do not substitute temperature for a temperature difference in E = mcΔT.

For a monatomic ideal gas, absolute temperature is proportional to mean translational kinetic energy: $E_k = \tfrac32 k_B T$. Internal energy is the total intermolecular potential energy plus the total random kinetic energy of the molecules; a phase change changes particle behaviour at constant temperature. Heating quantities:

Worked check. Melting then warming 0.40 kg of ice at 0 °C to 30 °C ($c = 4186$, $L_f = 3.34\times10^5$ J/kg): $Q_1 = 0.40 \times 3.34\times10^5 = 1.34\times10^5$ J; $Q_2 = 0.40 \times 4186 \times 30 = 5.0\times10^4$ J; total $1.84\times10^5$ J.

Conduction transfers energy down a temperature gradient at a rate set by the material, area and gradient:

Worked check. A window ($k = 0.84$, area 1.5 m², thickness 4.0 mm) with a 20 K inside–outside difference: $\Delta Q/\Delta t = 0.84 \times 1.5 \times 20/0.0040 = 6.3\times10^3$ W.

Radiation from a black body follows the Stefan–Boltzmann law $L = \sigma A T^4$; the apparent brightness falls with distance as $b = L/(4\pi d^2)$, and Wien's displacement law $\lambda_\text{max}T = 2.9\times10^{-3}\ \mathrm{m\,K}$ locates the peak wavelength.

Worked check (Wien). The Sun's spectrum peaks at 580 nm: $T = 2.9\times10^{-3}/5.8\times10^{-7} = 5.0\times10^3$ K.

Worked check (Stefan–Boltzmann). A person (area 1.8 m², T = 303 K, $\sigma = 5.7\times10^{-8}$): $L = 5.7\times10^{-8} \times 1.8 \times 303^4 \approx 8.7\times10^2$ W radiated.

Emissivity compares a real surface with a black body at the same temperature: emissivity = (power radiated per unit area)$/\sigma T^4$. The guide's guidance notes: Kelvin–Celsius conversion is required, and $\Delta T$ is identical on both scales.

Key:

Thermal measurements and particle models: Temperature relates to particle motion in a model; internal energy includes kinetic and potential contributions. During a change of state, energy can change particle arrangements rather than temperature.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• One cold day does not disprove a warming climate. Weather describes short-term conditions; climate describes distributions over longer times and regions.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• The Earth energy balance includes incoming solar radiation, reflection, absorption and outgoing infrared radiation. Greenhouse gases absorb and emit infrared radiation. Feedback can alter the response to an initial forcing.
• mitigation 减缓: Actions addressing causes of environmental change; adaptation 适应: Actions reducing harm from environmental impacts.

Choose evidence that can test it

• Distinguish mitigation, which addresses drivers, from adaptation, which reduces harm from impacts. A policy assessment needs evidence about effectiveness, cost, equity and uncertainty; one criterion is not the entire decision.
• Compare multi-year data using consistent baselines. State the region, timescale and uncertainty. At HL, connect a management decision to law, economics and ethics rather than treating these lenses as extra definitions only.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a surface receives 200 power units and reflects 50. Absorbed input = incoming-reflected = 200-50 = 150. Reflected fraction = 50/200 = 0.25 = 25%. A change in reflectivity alters the absorbed budget under this model.

Example:

Incoming energy is 240 units and reflected energy 72. Find reflected percentage. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The greenhouse effect is not the same process as ozone depletion. A carbon footprint estimate depends on its system boundary.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

One local weather observation establishes a global climate trend. This claim is false: The greenhouse effect is not the same process as ozone depletion. A carbon footprint estimate depends on its system boundary.

Emissivity compares a surface's radiated power per unit area with a black body's $\sigma T^4$; albedo is the fraction of incident power scattered (reflected):

Earth's albedo varies daily with cloud and latitude. The solar constant S is the incoming intensity; averaged over the sphere, the mean incoming intensity is $S/4$ (a planet presents a disc $\pi R^2$ but spreads the power over $4\pi R^2$). Equilibrium comes from energy balance between incoming and outgoing radiation — the guide's stated problem type:

Worked check (equilibrium temperature). With $S = 1361\ \mathrm{W/m^2}$ and albedo 0.30, absorbed mean intensity $= 0.70 \times 1361/4 = 238\ \mathrm{W/m^2}$. Setting emission equal: $\sigma T^4 = 238$, so $T = (238/5.7\times10^{-8})^{1/4} = 255$ K — the effective radiating temperature; the surface is warmer because of the greenhouse effect.

The greenhouse gases — methane CH₄, water vapour, carbon dioxide CO₂ and nitrous oxide N₂O — have natural and human origins. Each absorbs infrared through transitions between molecular energy levels, then re-emits in all directions, returning part of the outgoing radiation to the surface. The effect can be described by a resonance model or by molecular energy levels; human enhancement of it is the enhanced greenhouse effect.

Key:

Climate evidence, energy budgets and policy: Distinguish mitigation, which addresses drivers, from adaptation, which reduces harm from impacts. A policy assessment needs evidence about effectiveness, cost, equity and uncertainty; one criterion is not the entire decision.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A sealed gas container changes pressure when heated. Celsius ratios cannot predict the pressure change because the gas model uses absolute temperature.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• In a gas model, particles move randomly and pressure results from momentum transfer at walls. The ideal gas 理想气体 equation connects pressure, volume, amount and absolute temperature.
• absolute temperature: Temperature on the kelvin scale; ideal gas: A gas model with specified simplifying assumptions.

Choose evidence that can test it

• At fixed amount and volume, pressure is proportional to kelvin temperature. At fixed temperature and amount, pressure is inversely proportional to volume. State which quantities are fixed before choosing a relationship.
• Use approved apparatus with a temperature range and pressure limit set by the teacher. Allow thermal equilibrium and record pressure against kelvin temperature. Never heat an improvised sealed vessel.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: pressure is 100 kPa at 300 K, with fixed volume and amount. At 330 K, p2/p1=T2/T1. p2=p1 T2/T1=100×330/300=110 kPa. A 30 °C rise is a 30 K change, but the temperature ratio must use kelvin.

Example:

At fixed volume, pressure is 120 kPa at 300 K. Find pressure at 350 K. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• An ideal gas is a model with conditions of validity. Celsius zero is not zero molecular motion, and internal energy is not determined by pressure alone.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A gas pressure ratio can always use Celsius temperatures. This claim is false: An ideal gas is a model with conditions of validity. Celsius zero is not zero molecular motion, and internal energy is not determined by pressure alone.

Pressure is force per area; the amount of substance counts molecules: $n = N/N_A$. The ideal gas is a modelled system that approximates a real gas; its law follows from the empirical gas laws:

Molecular collisions give pressure, relating it to the mean molecular speed: $P = \tfrac13 \rho \langle v^2\rangle$. The internal energy of an ideal monatomic gas is purely kinetic:

Worked check. A 2.0 mol sample at 300 K, 1.0 atm ($1.01\times10^5$ Pa): $V = nRT/P = 2.0 \times 8.31 \times 300/1.01\times10^5 = 4.9\times10^{-2}\ \mathrm{m^3}$; $U = \tfrac32 \times 2.0 \times 8.31 \times 300 = 7.5\times10^3$ J.

Changes of state can be drawn on pressure–volume diagrams (isobaric, isochoric, isothermal lines). An ideal gas approximates a real gas best at high temperature and low pressure (low density), far from condensation; near liquefaction the molecules' own volume and attractions matter.

Key:

Gas models and absolute temperature: At fixed amount and volume, pressure is proportional to kelvin temperature. At fixed temperature and amount, pressure is inversely proportional to volume. State which quantities are fixed before choosing a relationship.

Full B.4 Thermodynamics coverage (HL ONLY — the guide states there is no standard-level content in B.4), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 43–45). Clausius sign convention (Q to the system, W by the system); closed vs isolated systems distinguished.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Compressing a gas can warm it even without a heater. Work is another path for energy transfer into the system.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Internal energy 内能 is a property of the chosen system. Heat and work describe energy transfer across its boundary. For an ideal monatomic gas, internal energy depends on temperature and amount.
• adiabatic 绝热的: A process with no heat transfer; internal energy: Total microscopic kinetic and potential energy of a system.

Choose evidence that can test it

• State a sign convention before using the first law. Here change in internal energy equals heat into the system minus work done by the system. Expansion work can reduce internal energy when no heat enters.
• Identify whether a process is isothermal, adiabatic or at constant volume. On a pressure-volume graph, area under the process curve gives work done by the gas. The path matters for work even when endpoints match.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: gas gains 300 J by heating and does 120 J of work. With ΔU = Q-W, ΔU = 300-120 = 180 J. For adiabatic compression, Q=0 and work done by the gas is negative, so ΔU is positive.

Example:

A gas gains 500 J of heat and does 200 J work. Find internal-energy change. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Adiabatic does not mean constant temperature. Use the stated sign convention; some courses write work done on the gas with the opposite sign.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every adiabatic process has constant temperature. This claim is false: Adiabatic does not mean constant temperature. Use the stated sign convention; some courses write work done on the gas with the opposite sign.

The first law applies conservation of energy to a closed system (Clausius sign convention: Q supplied to the system, W done by the system):

Entropy describes the number of accessible microstates: for reversible heat transfer at constant temperature $\Delta S = Q_\mathrm{rev}/T$, microscopically $S = k_B \ln\Omega$ (Ω = number of microstates). The second law sets the direction of real processes: the entropy of an isolated system never decreases; real processes are almost always irreversible. A non-isolated system's entropy can fall locally, compensated by a greater rise in the surroundings.

The four named processes keep one variable fixed — isovolumetric, isobaric, isothermal, adiabatic (for a monatomic ideal gas, $PV^{5/3} = \text{constant}$). Cyclic processes run heat engines, whose efficiency is $\eta = \text{useful work}/\text{input energy}$, limited by the Carnot cycle:

Worked check. An ideal monatomic gas is heated at constant volume: 500 J supplied, W = 0, so $\Delta U = 500$ J and $\Delta T = 500/(1.5 \times 2.0 \times 8.31) = 20$ K for 2.0 mol.

Worked check (Carnot). A reservoir pair at 900 K and 300 K: $\eta_\text{Carnot} = 1 - 300/900 = 0.67$ — no engine between these temperatures can beat 67 %.

Worked check (entropy). 1200 J transferred from a large reservoir at 300 K: $\Delta S = 1200/300 = 4.0$ J/K.

Key:

Thermodynamic systems and energy accounts: State a sign convention before using the first law. Here change in internal energy equals heat into the system minus work done by the system. Expansion work can reduce internal energy when no heat enters.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A lamp becomes dimmer when another is added in series. Current and energy transfer depend on the whole circuit, not only one lamp.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Current is charge passing per time. Potential difference is energy transferred per charge. Resistance is potential difference divided by current for a stated operating point.
• current: Rate of flow of charge; potential difference: Energy transferred per unit charge.

Choose evidence that can test it

• Current is the same through components in series. Potential differences add around the series path. In parallel, branches share the same potential difference, while branch currents sum at a junction.
• Place an ammeter in series and a voltmeter in parallel. For an I-V investigation, change voltage in steps, reverse polarity when appropriate and limit current to reduce heating.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a resistor has 6.0 V across it and carries 0.30 A. Use R = V/I. R = 6.0/0.30 = 20 ohms. Power = VI. Power = 6.0×0.30 = 1.8 W. These quantities describe the same operating point.

Example:

A component has 12 V across it and carries 0.40 A. Find resistance. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A lamp uses up electric current as charge passes through it. This claim is false: Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.

A cell is a source of emf $\varepsilon$ — the energy it gives each coulomb of charge. Chemical cells convert stored chemical energy; solar cells convert radiant energy. Comparing sources is part of the guide: solar cells are renewable and need no fuel but work only in daylight and their manufacture costs energy; chemical cells are portable and reliable anywhere but store a finite amount of energy and become waste.

The full B.5 equation set:

where $r$ is the cell's internal resistance: the terminal pd is $V = \varepsilon - Ir = IR$, lower than the emf whenever current flows.

Resistivity $\rho$ is the material property behind a wire's resistance: $R = \rho L / A$, so long thin wires resist more, short thick ones less. Conduction is decided by the mobility of charge carriers, not their mere presence: a metal's free electrons move easily; an insulator's charges are locked in place.

Ohm's law: at constant temperature a metal conductor's I–V graph is a straight line through the origin — ohmic behaviour. A filament lamp heats as it conducts, so its curve bends upward in V — non-ohmic; a diode conducts one way only. Circuit symbols come from the Physics data booklet; an ideal ammeter (in series) has zero resistance, an ideal voltmeter (in parallel) infinite resistance.

Variable resistors: a thermistor has resistance that falls as temperature rises; an LDR has resistance that falls as light intensity rises; a potentiometer divides a pd through a sliding contact.

Worked check (emf and terminal pd). A cell of emf $1.5\ \mathrm{V}$ and internal resistance $0.50\ \Omega$ drives a $7.5\ \Omega$ resistor: $I = 1.5/(7.5 + 0.50) = 0.19\ \mathrm{A}$; terminal pd $= 1.5 - 0.19 \times 0.50 = 1.4\ \mathrm{V}$, which equals $IR = 0.19 \times 7.5 = 1.4\ \mathrm{V}$. The cell delivers $P = \varepsilon I = 0.28\ \mathrm{W}$, of which $I^2 r = 1.8\times10^{-2}\ \mathrm{W}$ heats the cell itself.

Worked check (resistivity). Copper has $\rho = 1.7\times10^{-8}\ \Omega\,\mathrm{m}$. A wire of length 1.5 m and cross-section $0.20\ \mathrm{mm^2} = 2.0\times10^{-7}\ \mathrm{m^2}$ has $R = \rho L/A = (1.7\times10^{-8} \times 1.5)/(2.0\times10^{-7}) = 0.13\ \Omega$.

Worked check (series–parallel). $6.0\ \Omega$ and $3.0\ \Omega$ in parallel: $1/R_p = 1/6.0 + 1/3.0 = 0.50$, so $R_p = 2.0\ \Omega$; in series they give $9.0\ \Omega$. Across a $6.0\ \mathrm{V}$ ideal battery the parallel pair draws $I = 6.0/2.0 = 3.0\ \mathrm{A}$, splitting into $1.0\ \mathrm{A}$ and $2.0\ \mathrm{A}$ — the smaller resistor carries the larger share.

Key:

Cells, resistivity and internal resistance: series shares one current and adds potential differences; parallel shares one potential difference and adds currents. The emf drives the whole loop resistance including the cell's own internal resistance, and the terminal pd falls below the emf by exactly $Ir$.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A swing gains a large amplitude when pushes arrive with the right timing. The forcing frequency and damping 阻尼 help determine the response.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Simple harmonic motion has acceleration proportional to displacement and directed toward equilibrium. Period is time per cycle; frequency is its reciprocal. Damping transfers energy away from the oscillating system.
• resonance: A large response to periodic forcing near a natural frequency; damping: Energy transfer out of an oscillating system.

Choose evidence that can test it

• Velocity is greatest near equilibrium for ideal SHM, while acceleration magnitude is greatest at extreme displacement. Resonance can occur near the natural frequency under periodic driving, with amplitude limited by damping.
• Measure time for several complete oscillations and divide. Define a cycle consistently and use a small displacement when the model requires it. Keep pendulum paths clear and record damping effects rather than assuming perfect motion.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: 20 cycles take 32 s. T=total time/cycles=32/20=1.6 s. Frequency=1/T=1/1.6=0.625 Hz. A shorter total timing interval would make reaction-time error a larger fraction of the measurement.

Example:

30 cycles take 45 s. Find period. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Not every repeated motion is SHM. A force proportional to displacement but directed away from equilibrium does not produce SHM.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every periodic motion satisfies the condition for simple harmonic motion. This claim is false: Not every repeated motion is SHM. A force proportional to displacement but directed away from equilibrium does not produce SHM.

Simple harmonic motion (SHM) needs a restoring force proportional to displacement, directed back toward the equilibrium position. The defining equation folds that into acceleration:

The minus sign is the physics: acceleration always points against the displacement, so the particle is always pulled home. One oscillation is described by its amplitude $x_0$, period $T$, frequency $f$ and angular frequency $\omega$, linked by:

An ideal mass–spring system gives SHM; a simple pendulum approximates SHM for small angles. Their periods are:

Energy swaps between kinetic and potential through the cycle (qualitatively): total energy constant, kinetic energy greatest as the particle passes equilibrium at maximum speed, potential energy greatest at the extremes where the particle momentarily stops.

Worked check (mass–spring). A 0.50 kg mass on a spring of stiffness $20\ \mathrm{N/m}$: $T = 2\pi\sqrt{0.50/20} = 2\pi(0.158) = 0.99\ \mathrm{s}$, so $f \approx 1.0\ \mathrm{Hz}$ and $\omega = 2\pi/0.99 = 6.3\ \mathrm{rad/s}$.

Worked check (pendulum). A pendulum of length 1.0 m ($g = 9.8\ \mathrm{m/s^2}$): $T = 2\pi\sqrt{1.0/9.8} = 2.0\ \mathrm{s}$, $f = 0.50\ \mathrm{Hz}$.

Worked check (maximum acceleration). Amplitude $0.050\ \mathrm{m}$, period $0.50\ \mathrm{s}$: $\omega = 2\pi/0.50 = 12.6\ \mathrm{rad/s}$, and $a_{max} = \omega^2 x_0 = 158 \times 0.050 = 7.9\ \mathrm{m/s^2}$, at the extremes.

Simple harmonic motion: the restoring condition $a = -\omega^2 x$ with $T = 1/f = 2\pi/\omega$; mass–spring $T = 2\pi\sqrt{m/k}$ and pendulum $T = 2\pi\sqrt{l/g}$; energy trades between kinetic (max at equilibrium) and potential (max at the extremes) while the total stays constant.

At higher level the motion and its energy become fully algebraic. With phase angle $\phi$ in radians:

Phase constant: $\phi = 0$ starts at equilibrium moving positive; $\phi = \pi/2$ starts at rest at $+x_0$.

Worked check (energy at a displacement). $m = 0.20\ \mathrm{kg}$, $x_0 = 0.040\ \mathrm{m}$, $\omega = 5.0\ \mathrm{rad/s}$: $E_T = \tfrac{1}{2}(0.20)(25)(0.040^2) = 4.0\times10^{-3}\ \mathrm{J}$. Halfway out at $x = 0.020\ \mathrm{m}$: $E_p = \tfrac{1}{2}(0.20)(25)(0.020^2) = 1.0\times10^{-3}\ \mathrm{J}$, leaving $E_k = 3.0\times10^{-3}\ \mathrm{J}$, and $v = \omega\sqrt{x_0^2 - x^2} = 5.0\sqrt{0.0012} = 0.17\ \mathrm{m/s}$ — consistent, since $\tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.20)(0.17^2) = 2.9\times10^{-3}\ \mathrm{J} \approx E_k$.

Key:

SHM at higher level: $x = x_0\sin(\omega t+\phi)$, $v = \omega x_0\cos(\omega t+\phi)$, $v = \pm\omega\sqrt{x_0^2-x^2}$, $E_T = \tfrac{1}{2}m\omega^2x_0^2$ and $E_p = \tfrac{1}{2}m\omega^2x^2$; kinetic energy is the gap between total and potential.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A wave can carry energy across water while a floating marker mainly oscillates. Energy transfer and bulk transfer of matter are different.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Frequency 频率 is oscillations per time; wavelength 波长 is distance between successive points in phase. Wave speed equals frequency multiplied by wavelength. Transverse oscillations are perpendicular to propagation.
• wavelength: Distance between successive points in phase; frequency: Number of oscillations per unit time.

Choose evidence that can test it

• At a boundary, frequency stays fixed by the source. A change of speed changes wavelength. Refraction follows from speed differences; angles are measured from the normal.
• Measure several wavelengths and divide to reduce fractional reading uncertainty. Use a ray box with a normal drawn at the boundary. Keep the beam away from eyes and record incident and refracted angles clearly.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: frequency 5 Hz and wavelength 0.40 m. Use v = fλ. v = 5×0.40 = 2.0 metres per second. If speed falls to 1.5 metres per second at the same frequency, wavelength = v/f = 1.5/5 = 0.30 m.

Example:

Frequency is 8 Hz and wavelength is 0.25 m. Find wave speed. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The wave frequency does not change simply because the medium changes. A longitudinal wave has oscillations parallel to propagation.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Wave frequency always changes when a wave enters a new medium. This claim is false: the frequency is set by the source and stays fixed; it is the speed and the wavelength that change together.

A travelling wave carries energy through a medium by local disturbance, with no net displacement of the medium itself. Two geometries:

• Transverse wave — particle oscillations perpendicular to propagation (waves on a rope, all electromagnetic waves).
• Longitudinal wave — oscillations parallel to propagation, as compressions and rarefactions (sound).

Particle motion is described two ways, and the guide requires both: displacement against position along the wave at one instant (a snapshot of the wave profile) and displacement against time at one fixed point (how a single particle oscillates as the wave passes).

Sound is mechanical and longitudinal: it needs a medium, and its speed depends on that medium (about $340\ \mathrm{m/s}$ in air, faster in water, faster still in steel). Electromagnetic waves are transverse oscillations of electric and magnetic fields, perpendicular to each other and to the direction of travel; they need no medium and move through vacuum at $c = 3.0\times10^8\ \mathrm{m/s}$. That is the mechanical-versus-electromagnetic divide: matter versus field, medium versus vacuum.

The electromagnetic spectrum runs from radio through microwave, infrared, visible (about $400$–$700\ \mathrm{nm}$), ultraviolet and X-rays to gamma rays, each band an order of magnitude or more in wavelength — the data booklet lists the approximate values.

Worked check (sound). A $170\ \mathrm{Hz}$ note in air ($v = 340\ \mathrm{m/s}$): $\lambda = v/f = 340/170 = 2.0\ \mathrm{m}$, and the period $T = 1/f = 5.9\times10^{-3}\ \mathrm{s}$.

Worked check (radio wave). An FM station at $100\ \mathrm{MHz}$: $\lambda = c/f = (3.0\times10^8)/(1.0\times10^8) = 3.0\ \mathrm{m}$ — a radio wavelength.

Worked check (change of medium). A water wave of frequency $2.0\ \mathrm{Hz}$ passes from deep water ($v = 0.60\ \mathrm{m/s}$, $\lambda = 0.30\ \mathrm{m}$) to shallow ($v = 0.40\ \mathrm{m/s}$). Frequency is set by the source and unchanged, so the wavelength falls to $\lambda = v/f = 0.40/2.0 = 0.20\ \mathrm{m}$.

Key:

Travelling waves: transverse oscillate across the propagation direction, longitudinal along it; $v = f\lambda = \lambda/T$ with frequency fixed by the source when the medium changes; sound is mechanical and longitudinal, electromagnetic waves are transverse fields that cross a vacuum at $c$.

Full guide C.3 scope (Wave phenomena) including the AHL block, first assessment 2025. Read directly from the acquired Physics guide, pages 47–48. Prerequisites: C.2 (v = fλ, transverse and longitudinal waves) and basic trigonometry.

These are original teaching examples, not actual measurements or completed assessed learner investigations.

In two and three dimensions a wave is drawn two ways: wavefronts — lines joining points in step (in phase), one wavelength apart — and rays — arrows perpendicular to the wavefronts showing the direction of travel. Sketches in this topic are limited to incident, reflected and transmitted waves.

At a boundary between media a wave does some or all of three things:

• Reflection — the wavefront returns from the boundary (angle of incidence = angle of reflection, both measured from the normal).
• Transmission — part of the wave passes into the second medium.
• Refraction — the transmitted wave changes direction because its speed changes: toward the normal entering a slower medium, away from it entering a faster one. Frequency is fixed by the source; wavelength changes with speed.

The refraction rule in ratio form:

where $n$ is the refractive index of each medium and $\theta$ the angle between the ray and the normal. Equivalently $n_1\sin\theta_1 = n_2\sin\theta_2$; the larger refractive index means the slower speed in that medium.

Travelling from a slower to a faster medium (glass to air, $n_1 > n_2$), there is an angle beyond which no ray can escape: at the critical angle the refracted ray runs along the boundary,

and beyond it all the light reflects totally back into the medium — total internal reflection (TIR), the principle of optical fibres.

Worked check (refraction into glass). Air to glass, $n_{air} = 1.0$, $n_{glass} = 1.5$, incidence $40°$: $\sin\theta_2 = (n_1/n_2)\sin\theta_1 = (1.0/1.5)(0.643) = 0.429$, so $\theta_2 = 25°$ — bent toward the normal, as expected entering a slower medium. The speeds follow the same ratio: $v_{glass} = c/1.5 = 2.0\times10^8\ \mathrm{m/s}$.

Worked check (critical angle). The same glass against air: $\sin\theta_c = 1.0/1.5 = 0.667$, so $\theta_c = 42°$. Any ray inside the glass hitting the wall at more than $42°$ to the normal is totally internally reflected.

A wave meeting an obstacle or an opening does not travel in a perfectly straight line: it bends and spreads — diffraction. The spreading is strongest when the aperture is comparable to or smaller than the wavelength, which is why sound bends around doorways (metre wavelengths) while light (hundreds of nanometres) appears to travel straight. A wavefront-ray diagram shows plane wavefronts entering a slit and leaving as curved, spreading fronts.

When two waves meet at a point, displacements add — superposition. For a steady pattern the sources must be coherent: same frequency and a constant phase difference. Then the outcome depends on the path difference from the two sources:

with $n$ an integer. Constructive interference builds a maximum (two crests arrive together); destructive cancels to a minimum (a crest meets a trough).

Worked check. Two coherent sources in phase, wavelength 0.50 m. A point 3.00 m from one source and 3.75 m from the other has path difference 0.75 m $= 1.5\lambda = (1 + \tfrac{1}{2})\lambda$ — destructive: a minimum. Moving the point until the path difference is 1.00 m $= 2\lambda$ restores a maximum.

Light through one narrow slit then two narrow slits (coherent by construction) produces bright and dark fringes on a screen — interference, the evidence that light is a wave. The fringe separation is

where $d$ is the slit separation and $D$ the slit-to-screen distance.

Worked check (measuring a wavelength). Slits 0.20 mm apart, screen 2.4 m away, and 12 bright fringes counted across 55 mm (11 gaps, so $s = 55/11 = 5.0\ \mathrm{mm}$): $\lambda = sd/D = (5.0\times10^{-3})(0.20\times10^{-3})/2.4 = 4.2\times10^{-7}\ \mathrm{m}$ — violet-blue light.

Single-slit diffraction: light of wavelength $\lambda$ through one rectangular slit of width $b$ (monochromatic, normal incidence) spreads into a broad central maximum with weaker side fringes. The first minimum sits at

so a narrower slit spreads the light wider — diffraction grows as the aperture shrinks toward the wavelength. The effect of slit width on the intensity is qualitative: the central peak dominates, and its width is what the equation tracks.

A real double-slit is two single slits, so the fine double-slit fringes ride under the broader single-slit pattern: the single-slit pattern modulates the double-slit interference pattern — fringe heights follow the envelope of one slit's diffraction.

Many parallel slits — a diffraction grating of spacing $d$ — sharpen each maximum into a bright narrow line at

with $n = 0, 1, 2, \ldots$ the order. Because each wavelength has its own angle, white light through a grating splits into a spectrum per order (violet nearest the straight-through direction), and the number of visible orders is capped by $\sin\theta \le 1$.

Worked check (single slit). $b = 0.10\ \mathrm{mm}$, $\lambda = 500\ \mathrm{nm}$: $\sin\theta = 5.0\times10^{-7}/1.0\times10^{-4} = 5.0\times10^{-3}$, so $\theta = 0.29°$. Widening the slit to $0.40\ \mathrm{mm}$ divides the angle by four: $\theta = 0.072°$ — the pattern contracts as the aperture grows.

Worked check (grating orders). A grating with 600 lines per millimetre: $d = 1.0\times10^{-3}/600 = 1.7\times10^{-6}\ \mathrm{m}$. With $\lambda = 600\ \mathrm{nm}$: order 1 at $\sin\theta = 0.36$ ($\theta = 21°$), order 2 at $\sin\theta = 0.72$ ($\theta = 46°$), and order 3 would need $\sin\theta = 1.08$ — impossible, so only two orders appear. At $400\ \mathrm{nm}$ (violet) order 1 sits at $\sin\theta = 0.24$ ($14°$): one grating, one flash of white light, and the first-order spectrum spreads from $14°$ to $21°$.

1. State what is meant by a wavefront and how a ray is related to it.
2. A ray crosses from air into water ($n = 1.3$) at $50°$ to the normal. Calculate the refraction angle.
3. State the condition for total internal reflection, and why a fish tank can mirror-image objects at glancing angles.
4. Compare the diffraction of sound and of light through a doorway, with a reason.
5. Two coherent sources 0.040 m apart emit 0.060 m waves in phase. A point lies 0.30 m from one source along a line perpendicular to the source line; state its path difference class (constructive or destructive) if it is 0.33 m from the other source.
6. In a double-slit set-up the slit separation is halved and the screen moved twice as far away. What happens to the fringe separation?
7. A double-slit experiment with green light gives fringes 4.0 mm apart. Estimate the new separation if blue light of half the wavelength is used with the same geometry.

Answers: 1 a line of points in phase, one wavelength apart; the ray is everywhere perpendicular to the wavefronts and shows travel direction; 2 $\sin\theta_2 = (1.0/1.3)\sin 50° = 0.588$, $\theta_2 = 36°$; 3 travel from a slower (higher-$n$) medium toward a faster one at more than the critical angle — then no transmission exists; beyond the critical angle the inside of the glass surface acts as a mirror; 4 sound (wavelengths of metres, comparable with the doorway) diffracts strongly around the opening, light (wavelengths hundreds of nanometres, far smaller) spreads negligibly and appears to travel straight; 5 path difference 0.03 m $= \tfrac{1}{2}(0.060\ \mathrm{m})$ — destructive; 6 $s = \lambda D/d$ doubles from $D$ and doubles again from $d$ — four times wider; 7 $s \propto \lambda$, so 2.0 mm.

Full guide C.4 scope (Standing waves and resonance), first assessment 2025. Read directly from the acquired Physics guide, pages 48–49. Prerequisites: C.1 (oscillation vocabulary) and C.3 (superposition).

These are original teaching examples, not actual measurements or completed assessed learner investigations.

Two identical waves travelling in opposite directions — usually a wave and its reflection — superpose into a standing wave: the pattern stays put while the medium oscillates inside it. Where the two always cancel there is a node (zero displacement); where they always reinforce there is an antinode (maximum amplitude). Two points between neighbouring nodes oscillate in phase; points on opposite sides of a node are in antiphase — unlike a travelling wave, the phase does not march along.

The ends decide which wavelengths fit. A fixed string end (or the closed end of a pipe) must be a displacement node; a free string end (or the open end of a pipe) must be an antinode. Pipes are described in displacement terms only (pressure nodes are not required, and no end corrections).

The lowest-frequency mode is the first harmonic (the terms "fundamental" and "overtone" are not used in this course). Determining the wavelength and frequency of the $n$th harmonic from the length and wave speed is a required skill.

Worked check (string). A string of length 0.60 m carries waves at 240 m/s. First harmonic: $\lambda_1 = 2L = 1.2\ \mathrm{m}$, $f_1 = v/2L = 240/1.2 = 200\ \mathrm{Hz}$; the second harmonic doubles to 400 Hz, the third to 600 Hz.

Worked check (closed–open pipe). A pipe closed at one end, open at the other, length 0.40 m, air at 340 m/s: $f_1 = v/4L = 340/1.6 = 213\ \mathrm{Hz}$; the next harmonic is the third, $f_3 = 3f_1 = 638\ \mathrm{Hz}$ — even harmonics cannot form.

Worked check (open–open pipe). The same pipe opened at both ends: $f_1 = v/2L = 340/0.80 = 425\ \mathrm{Hz}$, an octave above the closed–open first harmonic, and every harmonic is allowed.

Every elastic system has a natural frequency — the rate at which it oscillates when left alone. Drive it with a periodic force and the response depends on the driving frequency: when the driving frequency approaches the natural frequency the amplitude grows sharply — resonance. Energy arrives at exactly the rate the system can absorb it, so the amplitude at resonance is limited only by how fast energy dissipates.

Damping — the removal of oscillation energy — reshapes the response: it lowers the maximum amplitude, shifts the resonant peak slightly below the natural frequency, and broadens it.

• Light damping — a tall, sharp peak; large resonant amplitude (a swing pushed gently at its natural frequency).
• Critical damping — the oscillation dies away in the shortest time without passing the equilibrium point; the resonance peak has effectively vanished (car suspension).
• Heavy damping — slow, sluggish return; the system barely responds at any driving frequency (a heavy door closer).

Resonance is wanted and engineered (musical instruments, radio tuning, MRI) and unwanted or destructive (the Tacoma Narrows bridge, machinery shaking itself apart, soldiers breaking step on a bridge) — the guide asks for both sides.

1. State how a standing wave forms, and why nodes stay fixed while the medium between them moves.
2. Two points on a standing wave lie either side of a node, each 0.15 m from it. State their phase difference.
3. A string fixed at both ends is 0.75 m long and carries waves at 150 m/s. Calculate the first and second harmonic frequencies.
4. Explain why a closed–open pipe produces only odd harmonics.
5. A pipe of length 0.50 m is open at both ends; the air speed is 340 m/s. Calculate the first harmonic frequency and the wavelength of the second harmonic.
6. Sketch or describe how the resonance curve of a lightly damped system changes as damping increases.
7. Give one useful and one destructive effect of resonance, with the physical reason each happens.

Answers: 1 two identical waves travelling in opposite directions superpose; at a node their displacements always cancel, so it never moves, while between nodes the superposition oscillates; 2 antiphase — 180° (π rad) apart; 3 $f_1 = v/2L = 150/1.5 = 100\ \mathrm{Hz}$, $f_2 = 200\ \mathrm{Hz}$; 4 the closed end must be a node and the open end an antinode, so only an odd number of quarter-wavelengths fits — even harmonics cannot meet both conditions; 5 $f_1 = v/2L = 340/1.0 = 340\ \mathrm{Hz}$; $\lambda_2 = L = 0.50\ \mathrm{m}$; 6 the peak falls, broadens, and shifts slightly below the natural frequency; 7 useful — a radio circuit tuned to resonance picks one station out of many because it responds only near its natural frequency; destructive — driving at the natural frequency feeds energy in faster than it dissipates, so the amplitude grows until the structure fails.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A recognizable spectral line appears at a longer wavelength in a distant source. Its shift can provide evidence of recession along the line of sight.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Relative source-observer motion changes observed frequency or wavelength. For light at a relative speed much smaller than c, the magnitude of the fractional wavelength shift is approximately v/c. Recession corresponds to longer wavelength, and approach to shorter wavelength.
• redshift 红移: Shift of an identified line toward longer wavelength; rest wavelength 静止波长: Wavelength of the line measured with no relative source motion.

Choose evidence that can test it

• Compare a identified line with its laboratory rest wavelength. Use z = (observed−rest)/rest, then v approximately cz in the stated low-speed model. Do not apply this approximation without checking the regime or treating a cosmological redshift as a simple exact velocity.
• Use attributed spectra with calibration and line identification. Compare several lines for a consistent shift. State uncertainty and distinguish relative line-of-sight motion from an unmeasured transverse component.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 500 nm rest line is observed at 501 nm. z = (501−500)/500 = 0.002. With c=3.0×10⁸ m/s, v approximately 6.0×10⁵ m/s away. The sound-source denominator formula is not the required SL light approximation.

Example:

A 600 nm rest line is observed at 601.2 nm. Find z=(observed−rest)/rest. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The approximate shift relation does not show that light travels faster from an approaching source. A single unidentified line cannot safely establish a redshift.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

An observed redshift means light speed increased. This claim is false: The approximate shift relation does not show that light travels faster from an approaching source. A single unidentified line cannot safely establish a redshift.

Key:

Spectral shifts: a low-speed astronomical estimate: Compare a identified line with its laboratory rest wavelength. Use z = (observed−rest)/rest, then v approximately cz in the stated low-speed model. Do not apply this approximation without checking the regime or treating a cosmological redshift as a simple exact velocity.

What would explain this observation?

• An approaching sound source is heard at a higher pitch even when the source produces a constant frequency.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Doppler shift 多普勒频移 arises from relative source-observer motion. A moving source compresses wavefront 波面 spacing ahead and increases it behind. For sound in a stationary medium, the wave speed relative to that medium need not change.
• Doppler shift: Change of observed frequency due to relative motion; wavefront: A surface joining points with the same wave phase.

Choose evidence that can test it

• For a source approaching a stationary observer along the line of sight, f observed = f source × v/(v−v source). Recession uses v+v source in the denominator. These are sound-wave models; light requires its appropriate relation.
• Use a simulation or recorded data with stated source and observer directions. Draw wavefronts before choosing signs. Avoid demonstrations near moving traffic or unsafe high-volume sources.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: source frequency 500 Hz, sound speed 340 m/s, source approaches at 20 m/s. Observed frequency = 500 × 340/(340−20) = 531.25 Hz. The increase is 31.25 Hz; it is not the source frequency changing.

Example:

A 600 Hz source approaches a stationary observer at 20 m/s. Use v = 340 m/s. Find observed frequency. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A stationary source and moving observer require the observer-motion relation. Do not use the same denominator formula for every arrangement or import a sound formula into relativity.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

The Doppler effect always means the wave speed in the medium changed. This claim is false: A stationary source and moving observer require the observer-motion relation. Do not use the same denominator formula for every arrangement or import a sound formula into relativity.

Doppler effect: relative motion between a wave source and an observer changes the observed frequency (and wavelength) while the source keeps emitting steadily. In a wavefront diagram the moving source chases the circles it emits: wavefronts bunch ahead of it and stretch behind it, so an observer ahead receives more fronts per second (higher frequency) and one behind receives fewer.

For sound, the effect is qualitative at this level: pitch rises as a siren approaches and falls as it passes and recedes — the classic slide as an ambulance goes by. A moving observer meets the same wavefronts at a different rate and hears the same kind of shift.

For light, when the relative speed $v$ is much smaller than $c$, the fractional shift is

Approaching compresses the light (shorter wavelength — blueshift); receding stretches it (redshift). Because each element emits and absorbs at known rest wavelengths, a spectrum taken from a star or galaxy can be matched line by line: the whole pattern shifted long means the source recedes, and the shift size gives the radial speed.

Worked check (a galaxy line). A hydrogen line with rest wavelength 656 nm is observed at 659 nm: $\Delta\lambda/\lambda = 3/656 = 4.6\times10^{-3}$, so $v = 0.0046c = 1.4\times10^6\ \mathrm{m/s}$ away from us — a redshift, hence recession.

The effect is put to work everywhere: radar speed guns time the frequency shift of reflected radio waves, medical Doppler ultrasound measures blood-flow speed from the shift of echoes, and astronomy reads galaxy redshifts — the larger the shift, the faster the recession.

At higher level the sound shifts become exact. With $v$ the speed of sound and $u$ the speed of source or observer (never both moving in one problem):

The upper signs belong to approach (denominator shrinks for a source driving into its own wavefronts; numerator grows for an observer running into the fronts) and the lower signs to recession. The two cases are not interchangeable — a source moving at $30\ \mathrm{m/s}$ past a stationary observer gives a different pitch than an observer moving at $30\ \mathrm{m/s}$ past a stationary source, because the source changes the wavelength in the medium while the observer only changes the rate of crossing wavefronts.

Worked check (moving source). A 400 Hz siren travels at $30\ \mathrm{m/s}$ toward a stationary listener ($v = 340\ \mathrm{m/s}$): $f' = 400 \times 340/(340 - 30) = 439\ \mathrm{Hz}$. Moving away: $f' = 400 \times 340/(340 + 30) = 368\ \mathrm{Hz}$.

Worked check (moving observer). The listener cycles at $30\ \mathrm{m/s}$ toward a stationary 400 Hz source: $f' = 400 \times (340 + 30)/340 = 435\ \mathrm{Hz}$ — close to, but not the same as, the moving-source case.

In medical physics, Doppler ultrasound compares the echo frequency with the emitted pulse to map blood-flow speeds and heart-valve motion; radar systems read vehicle or aircraft speed from the shift of the returned wave — the same physics, different wave.

Key:

Doppler effect: bunching ahead and stretching behind a moving source raise the observed frequency on approach and lower it on recession; for light at $v \ll c$, $|\Delta f|/f \approx |\Delta\lambda|/\lambda \approx |v|/c$ reads straight off spectral lines; at higher level the sound equations split by mover — $f' = fv/(v \mp u_s)$ for a moving source, $f' = f(v \pm u_o)/v$ for a moving observer — with radar and medical ultrasound as the working applications.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• An orbiting satellite is continuously falling while moving sideways. Being in orbit does not require gravity to vanish.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Gravitational field 引力场 strength is force per unit mass. A gravitational force can provide the centripetal force 向心力 for a circular orbit. Field and potential describe different quantities.
• gravitational field: A description of gravitational force per unit mass; centripetal force: Net force toward the centre of a curved path.

Choose evidence that can test it

• For a point mass or outside a spherical mass, field strength follows an inverse-square distance dependence. Use distance from the centre, not height above the surface alone.
• State the circular-orbit approximation and ignore atmospheric drag only when justified. Draw the force toward the central body and velocity tangential to the orbit. Do not add an outward force merely because the path is circular.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: field strength is 8 units at distance r from a spherical source. At 2r, g2/g1=(r/2r)²=1/4. g2=8/4=2 units. The gravitational force on a fixed test mass decreases in the same ratio.

Example:

Field strength is 18 units at r. Find it at 3r for an inverse-square field. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Weightlessness in an orbiting spacecraft means apparent weight is small in free fall, not that there is no gravitational field.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Gravity is zero everywhere a spacecraft experiences apparent weightlessness. This claim is false: Weightlessness in an orbiting spacecraft means apparent weight is small in free fall, not that there is no gravitational field.

Newton's universal law of gravitation treats masses as points:

An extended body acts as a point mass when it is spherically symmetric (uniform density spheres behave as if all mass sat at the centre) or when the separation hugely exceeds its size. The gravitational field strength at a point is the force per unit mass on a small test mass there:

Gravitational field lines point inward toward the mass, never cross, and crowd together where the field is stronger: a radial pattern around a sphere, near-parallel lines close to a large surface. Where two bodies act, consider resultant-field calculations at points on the line joining them — the individual fields add as vectors, and somewhere between the bodies they cancel exactly.

Kepler's three laws of orbital motion:

1. Orbits are ellipses with the central body at one focus (a circle is the special case used in calculations).
2. A line from the central body to the orbiting body sweeps equal areas in equal times — faster when closer, slower when farther.
3. $T^2 \propto r^3$: the square of the period scales with the cube of the orbital radius.

Worked check (surface gravity). Earth: $M = 6.0\times10^{24}\ \mathrm{kg}$, $r = 6.4\times10^6\ \mathrm{m}$: $g = GM/r^2 = (6.67\times10^{-11} \times 6.0\times10^{24})/(6.4\times10^6)^2 = 9.8\ \mathrm{m/s^2}$. At twice that distance from the centre the inverse square gives $g = 9.8/4 = 2.5\ \mathrm{m/s^2}$.

Worked check (Kepler's third law). Mars orbits at 1.52 AU: $T = 1.52^{3/2} = 1.87$ Earth years — cube the radius ratio, take the square root.

Key:

Kepler and Newton: every pair of masses pulls with $F = Gm_1m_2/r^2$, giving $g = GM/r^2$ around a point or spherical mass; field lines run inward and crowd where the field strengthens; Kepler's laws govern the orbits — ellipses, equal areas in equal times, and $T^2 \propto r^3$.

Gravitational potential $V_g$ is zero at infinity and negative everywhere else — work done per unit mass bringing a mass in from infinity:

The field strength is the potential gradient: potential rises toward infinity, so $g$ points down the gradient. Equipotential surfaces (spheres around a lone mass) sit everywhere perpendicular to the field lines — a mass moved along one needs no work, and where equipotentials crowd the field is strong. Mapping a field with potential is a required skill. Two speeds anchor orbit problems:

Escape asks for total energy $\ge 0$: kinetic lifted to exactly $+GMm/r$ cancels the potential well, at any radius. A satellite's total orbital energy is $-GMm/2r$, so raising an orbit adds energy (potential up, kinetic down, total less negative). A small atmospheric drag force, qualitatively, drains orbital energy: the height decays, and because $v = \sqrt{GM/r}$ the orbital speed actually rises as the orbit shrinks.

Worked check (surface potential and escape). Earth: $V_g = -GM/r = -(4.0\times10^{14})/(6.4\times10^6) = -6.3\times10^7\ \mathrm{J/kg}$. Escape speed from the surface: $v_{esc} = \sqrt{2GM/r} = \sqrt{1.25\times10^8} = 1.1\times10^4\ \mathrm{m/s}$.

Worked check (orbit change). A $1000\ \mathrm{kg}$ satellite lifts its circular orbit from $r = 1.28\times10^7$ to $1.92\times10^7\ \mathrm{m}$ (2 to 3 Earth radii). Energy change $= \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = (2.0\times10^{17})(2.6\times10^{-8}) = 5.2\times10^9\ \mathrm{J}$ — the work the engines must supply.

Key:

Potential and orbits at higher level: $V_g = -GM/r$ with zero at infinity, $g$ its gradient, equipotentials perpendicular to field lines; $v_{esc} = \sqrt{2GM/r} = \sqrt{2}v_{orbital}$, a bound orbit carries $-GMm/2r$, drag decays the height while the orbital speed rises.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A negatively charged particle moves opposite to the direction of the electric field. Field direction is defined using a positive test charge.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Electric field strength 电场强度 is force per unit positive test charge. Like charges repel and unlike charges attract. Field lines point away from positive charges and toward negative charges. Closely spaced lines represent a stronger field within a consistently drawn diagram.
• electric field strength: Electric force per unit positive test charge; test charge: A small charge used to probe an electric field.

Choose evidence that can test it

• In a uniform field between ideal parallel plates, E = V/d away from edge effects. The force is F = qE, so its direction reverses for a negative charge. Coulomb force between ideal point charges decreases with separation squared.
• Sketch labelled plate polarities, field arrows and the charge before calculating. Convert separation to metres. In school, use simulations or approved low-voltage electrostatic models rather than exposed high-voltage equipment.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: plate potential difference is 120 V and separation is 0.020 m. E = V/d = 6,000 V/m. A +2.0 microcoulomb charge experiences F = qE = 0.012 N along the field; a −2.0 microcoulomb charge experiences the same magnitude opposite to it.

Example:

Use E = V/d. Find E for 100 V across 0.025 m. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Field strength is a vector. Potential difference is energy per charge and is not the same quantity. A field-line drawing is a model, not a set of physical threads.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every charged particle experiences a force along the electric field direction. This claim is false: only a positive charge is pushed along the field — a negative charge feels the force opposite to the field direction. A field-line drawing is a model, not a set of physical threads.

Coulomb's law mirrors gravitation with charge in place of mass — but with two signs:

Like charges repel, unlike attract. Charge is conserved — it moves, never appears or vanishes — and quantized: Millikan's oil-drop experiment (weight balanced by an electric force, $qE = mg$) found every drop's charge a whole-number multiple of $e = 1.6\times10^{-19}\ \mathrm{C}$, the electron charge.

Charge moves between bodies by friction (rubbing transfers electrons), by contact (shared on touching), and by induction — a charged body brought near a conductor separates its charges, and grounding (earthing) lets one sign drain away so the conductor is left with the opposite sign, without the charged body ever touching it.

The electric field strength is force per unit positive test charge, $E = F/q$, with field lines running from positive to negative, never crossing, denser where the field is stronger: radial around a point charge and outside a sphere, straight and equally spaced between opposite parallel plates where

Magnetic field lines close on themselves (no monopoles): a bar magnet runs N-out S-in; a straight wire carries concentric circles whose direction follows the right-hand grip rule (thumb along the current); a circular coil loops through its centre; a solenoid's interior is near-uniform, its outside like a bar magnet's.

Worked check (Coulomb). Two $+1.0\ \mu\mathrm{C}$ charges $0.050\ \mathrm{m}$ apart: $F = (9.0\times10^9 \times 1.0\times10^{-12})/(0.050)^2 = 3.6\ \mathrm{N}$ — repulsion, falling to one quarter in magnitude per doubling of separation.

Worked check (parallel plates). $12\ \mathrm{V}$ across $4.0\ \mathrm{mm}$: $E = V/d = 3000\ \mathrm{V/m}$, and the force on an electron between them is $F = eE = 4.8\times10^{-16}\ \mathrm{N}$.

Worked check (Millikan). A drop of mass $3.2\times10^{-15}\ \mathrm{kg}$ carrying three excess electrons floats when $qE = mg$: $E = mg/q = (3.2\times10^{-15} \times 9.8)/(4.8\times10^{-19}) = 6.5\times10^4\ \mathrm{V/m}$ — the whole-number charge pattern across many drops is the quantization evidence.

Key:

Electric and magnetic fields: Coulomb's $F = kq_1q_2/r^2$ with conserved, quantized charge (Millikan); $E = F/q$ read from field-line density, uniform $E = V/d$ between plates; charge moves by friction, contact or induction with grounding; magnetic field lines close on themselves — wire circles by the right-hand rule, solenoid like a bar magnet.

Electric potential is a scalar, zero defined at infinity, with the sign carried by the charge:

The field strength is the potential gradient, pointing down the slope of $V_e$. Equipotential surfaces sit everywhere perpendicular to the field lines: no work is done moving along one ($\Delta V_e = 0$). Recognize them as spheres around a point charge (and superposed shapes for up to four charges), a constant potential inside and just outside a charged conducting sphere where the interior field is zero, and planes between parallel plates. Work can be quoted in joules or electronvolts — the work done on one electron through one volt: $1\ \mathrm{eV} = 1.6\times10^{-19}\ \mathrm{J}$.

Worked check (potential of a point charge). At $0.30\ \mathrm{m}$ from a $+2.0\ \mu\mathrm{C}$ charge: $V_e = kQ/r = (9.0\times10^9 \times 2.0\times10^{-6})/0.30 = 6.0\times10^4\ \mathrm{V}$ — positive, with zero at infinity.

Worked check (electron accelerated). An electron crosses $500\ \mathrm{V}$: $W = q\Delta V_e = 500\ \mathrm{eV} = 8.0\times10^{-17}\ \mathrm{J}$, gaining the same kinetic energy from rest.

Worked check (charged sphere). A conducting sphere of radius $0.10\ \mathrm{m}$ carrying $+1.0\ \mu\mathrm{C}$: at the surface $V_e = kQ/r = 9.0\times10^4\ \mathrm{V}$; inside, charges sit on the surface, $E = 0$ and the potential stays at that single value — the sphere's interior is one equipotential volume.

Key:

Potential at higher level: $V_e = kQ/r$ (scalar, zero at infinity) with $E$ as its gradient and $W = q\Delta V_e$ in joules or electronvolts; equipotentials sit perpendicular to field lines, no work along them — spheres around points, constant inside a conductor, planes between plates.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A charged particle curves in a uniform magnetic field. Its direction changes although the magnetic force 磁力 alone does not increase its kinetic energy.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• The magnetic force magnitude is F = |q|vB sinθ. It is perpendicular to velocity and magnetic field, with direction determined by charge sign. For perpendicular entry into a uniform field, it can provide the centripetal force for circular motion.
• magnetic force: Force due to a magnetic field on a moving charge or current; momentum 动量: Mass multiplied by velocity in the classical model.

Choose evidence that can test it

• Equate magnetic force with mv²/r to obtain r = mv/(|q|B) for perpendicular motion. A stronger field makes a smaller radius at fixed momentum. Parallel entry gives zero magnetic force in this model.
• Draw velocity, field and force as separate arrows using stated into/out-of-page conventions. Use supplied beam data or a simulation. Do not treat a current-carrying wire direction as identical to electron-motion direction.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a particle has m = 2.0×10⁻²⁷ kg, v = 3.0×10⁶ m/s, |q| = 1.0×10⁻¹⁹ C and B = 0.20 T. r = mv/(|q|B) = 0.30 m. The calculation assumes non-relativistic motion and perpendicular entry.

Example:

Use r = mv/(|q|B). For m=2×10⁻²⁷ kg, v=2×10⁶ m/s, |q|=1×10⁻¹⁹ C and B=0.20 T, find r. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A magnetic force can change momentum direction without doing work. Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A magnetic force alone continuously increases the kinetic energy of a charged particle. This claim is false: A magnetic force can change momentum direction without doing work. Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.

Three motions cover this topic. In a uniform electric field the force $qE$ is constant, so a charge shot across the plates follows a parabola — accelerated along the field lines, unchanged across them. In a uniform magnetic field the force

acts perpendicular to velocity, so it bends the path into a circle without changing speed — kinetic energy stays constant. The radius comes from circular motion:

so a measured path radius yields the charge-to-mass ratio $q/m = v/(rB)$ — how the nature of a particle is read from its track. With perpendicularly crossed fields ($E$ and $B$), the electric and magnetic forces can cancel for one particular speed, $qE = qvB$:

a velocity selector: only that speed passes straight.

Currents feel the field too:

Parallel wires carrying currents the same way attract; opposite ways repel.

Worked check (proton circle). A proton at $v = 2.0\times10^6\ \mathrm{m/s}$ in $B = 0.50\ \mathrm{T}$: $r = mv/qB = (1.67\times10^{-27} \times 2.0\times10^6)/(1.6\times10^{-19} \times 0.50) = 4.2\times10^{-2}\ \mathrm{m}$. Reading the same radius backward gives $q/m = v/(rB) = 9.5\times10^7\ \mathrm{C/kg}$ — the proton's signature.

Worked check (velocity selector). $E = 1.5\times10^4\ \mathrm{V/m}$ crossed with $B = 0.050\ \mathrm{T}$: only $v = E/B = 3.0\times10^5\ \mathrm{m/s}$ goes straight — faster particles bend magnetic-dominant, slower ones electric-dominant.

Worked check (wires). A $0.30\ \mathrm{m}$ wire carrying $5.0\ \mathrm{A}$ perpendicular to $B = 0.20\ \mathrm{T}$ feels $F = BIL = 0.30\ \mathrm{N}$. Two long wires $0.050\ \mathrm{m}$ apart carrying $10\ \mathrm{A}$ each the same way: $F/L = \mu_0 I_1I_2/2\pi r = 4.0\times10^{-4}\ \mathrm{N/m}$, attraction.

Key:

Charged motion and currents: a uniform electric field gives a parabola, a magnetic field a circle at constant kinetic energy with $r = mv/qB$ — the path radius measures $q/m$ — and crossed fields select $v = E/B$; a wire carries $F = BIL\sin\theta$, and parallel wires attract when their currents run the same way with $F/L = \mu_0I_1I_2/2\pi r$.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A magnet beside a wire does not always produce a current. An induced electromotive force depends on changing magnetic flux linkage.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• A field describes a force effect at positions in space. A current-carrying conductor in a magnetic field can experience a force. Electromagnetic induction occurs when magnetic flux linkage changes.
• induction: Creation of emf by changing flux linkage; transformer 变压器: A device transferring energy between coils through changing flux.

Choose evidence that can test it

• Changing field strength, coil area, orientation or relative motion can change flux linkage. Lenz law describes an induced effect opposing the change producing it, consistent with energy conservation.
• Use a coil and sensitive meter to compare magnet motion in each direction. Record that a stationary arrangement gives no induced signal. Use approved low-voltage supplies for motor demonstrations.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a transformer has 200 primary turns, 50 secondary turns and 12 V primary voltage. For an ideal transformer, Vs/Vp = Ns/Np. Vs = Vp Ns/Np = 12×50/200 = 3.0 V. Real transformers also dissipate energy.

Example:

An ideal transformer has Np=400, Ns=100 and Vp=20 V. Find Vs. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A transformer needs changing magnetic flux; a steady DC input does not provide continuous transformer action. Magnetic field direction is not automatically the direction of force.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A steady DC input produces continuous ideal transformer action. This claim is false: A transformer needs changing magnetic flux; a steady DC input does not provide continuous transformer action. Magnetic field direction is not automatically the direction of force.

Magnetic flux through a surface measures field through it:

with $\theta$ between the field and the normal to the surface. A time-changing flux induces an emf — Faraday's law:

The minus sign is Lenz's law: the induced emf drives a current whose own field opposes the change causing it. That opposition is energy conservation made visible — something must do work against the opposing force to keep the change going; without it, a generator would spin itself for free. Three standard producers: a time-varying field (transformer), a coil entering or leaving a field region, and a straight conductor sweeping across the field:

A coil rotating in a uniform field sweeps its flux $\Phi = BA\cos\omega t$ sinusoidally through zero twice per turn, so the emf varies sinusoidally with peak $\varepsilon_0 = NBA\omega$: spinning faster raises both the frequency and the peak of the output. Self-induction — a coil's own changing current inducing an emf in itself that opposes the change — is qualitative only. Generator output grows with stronger field, more turns, larger area and faster rotation.

Worked check (flux). $B = 0.20\ \mathrm{T}$ through $A = 0.050\ \mathrm{m^2}$ tilted $30°$ from the normal: $\Phi = 0.20 \times 0.050 \times \cos 30° = 8.7\times10^{-3}\ \mathrm{Wb}$.

Worked check (Faraday). That flux collapsing to zero in $0.10\ \mathrm{s}$ in one loop: $\varepsilon = 8.7\times10^{-3}/0.10 = 8.7\times10^{-2}\ \mathrm{V}$; a 50-turn coil multiplies it to $4.3\ \mathrm{V}$.

Worked check (rotating coil). $N = 100$, $B = 0.25\ \mathrm{T}$, $A = 0.020\ \mathrm{m^2}$, $\omega = 31.4\ \mathrm{rad/s}$ (5 turns per second): $\varepsilon_0 = NBA\omega = 15.7 \approx 16\ \mathrm{V}$ peak, sinusoidal at 5 Hz.

Key:

Induction (HL): flux $\Phi = BA\cos\theta$, Faraday $\varepsilon = -N\Delta\Phi/\Delta t$ with Lenz's minus sign the face of energy conservation, moving-conductor $\varepsilon = BLv$, and a rotating coil delivering a sinusoid of peak $NBA\omega$ — grow the output with $B$, $N$, $A$ or speed.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• An excited gas produces separate coloured lines rather than every wavelength. The pattern is evidence for discrete atomic energy differences.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Atoms have a small positive nucleus and electrons. Emission and absorption spectra arise from transitions between discrete energy levels. A photon energy equals the level difference and obeys E = hf.
• emission spectrum 发射光谱: Wavelength pattern of radiation emitted by a source; energy level 能级: An allowed energy state in a model.

Choose evidence that can test it

• An emitted photon corresponds to a transition to a lower energy level. Absorption requires a compatible energy difference. Rutherford scattering supported a small dense nucleus, but that experiment alone did not establish the complete quantum model.
• Read a labelled energy-level diagram before calculating. Keep joules and electronvolts distinct and use the given constants. Compare attributed spectra at a common wavelength scale and avoid looking at unsafe light sources.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: energy levels are −6.0 and −2.0 eV. A downward transition releases 4.0 eV = 6.4×10⁻¹⁹ J using 1 eV = 1.6×10⁻¹⁹ J. With h = 6.4×10⁻³⁴ J s for this rounded exercise, frequency is 1.0×10¹⁵ Hz.

Example:

Levels are −7.0 and −2.0 eV. Find the energy emitted in the downward transition. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Negative bound-state energies are relative to a chosen zero; they do not mean a negative photon energy is emitted. A larger downward energy difference gives higher frequency.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A negative bound-state energy means an emitted photon has negative energy. This claim is false: Negative bound-state energies are relative to a chosen zero; they do not mean a negative photon energy is emitted. A larger downward energy difference gives higher frequency.

The Geiger–Marsden–Rutherford experiment fired alpha particles at thin gold foil. Most passed straight through — the atom is mostly empty space. A few deflected through large angles, and rare ones bounced almost straight back — impossible unless nearly all the positive charge and mass sit in a tiny, dense nucleus. The observation forced the plum-pudding model out and the nuclear model in.

Atoms are written in nuclear notation $^{\,A}_{Z}X$: $A$ the nucleon number (protons + neutrons), $Z$ the proton number, $X$ the chemical symbol.

Electrons in an atom occupy discrete energy levels. A photon is emitted when an electron drops to a lower level, absorbed when one climbs:

with $E$ the energy difference between the two levels. The photon frequencies an element can emit or absorb are therefore a fingerprint — its emission and absorption spectra — and each line corresponds to one specific transition. Matching a spectrum against known patterns reveals chemical composition, including in stars whose light no probe will ever reach.

Worked check (a transition). An atom has levels at $-1.5\ \mathrm{eV}$ and $-3.4\ \mathrm{eV}$. A downward jump releases $E = 1.9\ \mathrm{eV} = 3.0\times10^{-19}\ \mathrm{J}$, so $f = E/h = (3.0\times10^{-19})/(6.6\times10^{-34}) = 4.6\times10^{14}\ \mathrm{Hz}$ — red light. This pair of levels absorbs the same frequency on the upward transition; other allowed level pairs can absorb other frequencies: the line pair in emission and absorption match.

Key:

The nuclear atom: alpha scattering revealed a tiny dense nucleus; $^{A}_{Z}X$ counts nucleons and protons; electrons sit in discrete levels, emitting or absorbing photons with $E = hf$ set by the level gap — and the line pattern of those transitions identifies each element, near or astronomically far.

Nuclear radius grows with the cube root of nucleon number:

Since mass also grows with $A$ while volume grows with $R^3 \propto A$, nuclear density is the same for every nucleus — matter packed to about $10^{17}\ \mathrm{kg/m^3}$.

In head-on scattering, the alpha's kinetic energy converts entirely to electric potential energy at the turning point — Rutherford's energy conservation:

At higher energies the alpha passes close enough to sense the nucleus's finite size, and the scattering deviates from the Rutherford formula — the deviation itself measures the nuclear radius.

The Bohr model binds hydrogen's electron to quantized energies and orbits:

Quantized angular momentum forces quantized orbits, and those orbits carry the discrete energies the spectra demand — the model's triumph and, with its electron "orbit" picture, its eventual failure.

Worked check (closest approach). A $5.0\ \mathrm{MeV}$ alpha heads straight at gold ($Z = 79$): $r_{min} = 2kZe^2/KE = (2 \times 9.0\times10^9 \times 79 \times (1.6\times10^{-19})^2)/(8.0\times10^{-13}) = 4.5\times10^{-14}\ \mathrm{m}$ — about 45 fm, well outside the $\sim 7\ \mathrm{fm}$ gold nucleus.

Worked check (Bohr transition). The $n = 2 \to 1$ jump: $\Delta E = 13.6(1 - \tfrac{1}{4}) = 10.2\ \mathrm{eV}$, giving $\lambda = hc/\Delta E = (1.99\times10^{-25})/(1.63\times10^{-18}) = 1.2\times10^{-7}\ \mathrm{m}$ — the brightest ultraviolet line of the Lyman series.

Key:

AHL structure: $R = R_0A^{1/3}$ makes every nucleus equally dense; closest approach trades kinetic energy for electric potential ($r_{min} = 2kZe^2/KE$) and high-energy deviations expose the finite nucleus; Bohr's $E_n = -13.6/n^2$ eV follows from $mvr = nh/2\pi$ and predicts hydrogen's lines.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Bright low-frequency light can fail to eject electrons while dim higher-frequency light succeeds. Photon 光子 energy depends on frequency, not intensity alone.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• A photon has energy proportional to frequency. The work function 逸出功 is the minimum energy needed to release an electron from a surface. Above threshold, excess photon energy can become electron kinetic energy.
• photon: A quantum of electromagnetic radiation; work function: Minimum surface energy needed to release an electron.

Choose evidence that can test it

• Use photon energy = Planck constant × frequency. Maximum kinetic energy = photon energy - work function. Increasing intensity at fixed frequency increases photon arrival rate, not individual photon energy.
• Read axes carefully on a stopping-potential or kinetic-energy graph. Identify threshold frequency from the zero-energy intercept. State the metal and experimental conditions because work function is material-specific.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: photon energy is 5.0 electronvolts and work function is 2.0 electronvolts. Maximum kinetic energy = photon energy - work function = 5.0 - 2.0 = 3.0 electronvolts. A 1.5 electronvolt photon cannot cause emission from that surface in this model.

Example:

Photon energy is 4.8 eV and work function 2.1 eV. Find maximum kinetic energy. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Electrons do not accumulate many sub-threshold photons in the elementary single-photon model. Do not confuse photon number with photon energy.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Doubling light intensity doubles each photon energy at fixed frequency. This claim is false: Electrons do not accumulate many sub-threshold photons in the elementary single-photon model. Do not confuse photon number with photon energy.

The photoelectric effect — light ejecting electrons from a metal — is the classic evidence for the particle nature of light. Three observations defeat the classical wave theory outright: emission is instantaneous even in dim light (waves would need time to accumulate energy); no electrons appear below a threshold frequency, however intense the light; and maximum electron energy depends on frequency, not intensity. Einstein's photon explains all three: light arrives in quanta of $E = hf$, and one quantum either has enough energy or does not.

where $\Phi$ is the metal's work function — the minimum energy to release an electron. At the threshold frequency $f_0 = \Phi/h$, $E_{max}$ just reaches zero.

Matter runs the argument backward: electrons diffract through crystal planes — a wave property — so particles have a de Broglie wavelength

and matter is wave–particle dual too. In a diffraction experiment the first minimum of the particle pattern sits exactly where the de Broglie wavelength predicts it — the description required by the guide.

Compton scattering adds the knock-out blow for the particle picture: X-rays scattering off electrons emerge with a longer wavelength, exactly as billiard-ball collisions demand energy loss per photon:

Worked check (threshold and emission). A metal with $\Phi = 2.3\ \mathrm{eV}$: $f_0 = \Phi/h = (3.7\times10^{-19})/(6.6\times10^{-34}) = 5.6\times10^{14}\ \mathrm{Hz}$ — no emission below it. At $f = 7.0\times10^{14}\ \mathrm{Hz}$ the photon carries $2.9\ \mathrm{eV}$, so $E_{max} = 2.9 - 2.3 = 0.6\ \mathrm{eV}$.

Worked check (de Broglie). An electron at $2.0\times10^6\ \mathrm{m/s}$: $\lambda = h/mv = (6.6\times10^{-34})/(9.1\times10^{-31} \times 2.0\times10^6) = 3.6\times10^{-10}\ \mathrm{m}$ — the spacing of atomic planes, which is why crystals diffract electrons.

Worked check (Compton). At $\theta = 90°$: $\Delta\lambda = h/m_ec = 2.4\times10^{-12}\ \mathrm{m}$ — the Compton wavelength; at $180°$ the shift doubles to $4.9\times10^{-12}\ \mathrm{m}$, the maximum.

Key:

Quantum physics (HL): $E_{max} = hf - \Phi$ with threshold $f_0 = \Phi/h$ (instantaneity, threshold and frequency-dependence sink the wave theory); matter diffracts with $\lambda = h/p$; Compton scattering shifts photon wavelengths by $\Delta\lambda = (h/m_ec)(1-\cos\theta)$ — light as particles, matter as waves.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A detector records counts even when the classroom source is removed. Background must be considered before attributing every count to the source.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Radioactive decay is spontaneous and random for an individual nucleus. Half-life 半衰期 describes the time for half the undecayed nuclei in a large population to decay, or for background-corrected activity to halve.
• half-life: Time for activity or undecayed population to halve; background radiation 本底辐射: Radiation measured apart from the investigated source.

Choose evidence that can test it

• Subtract background counts measured over the same time interval. Distinguish irradiation from contamination. Shielding, distance and reduced exposure time can reduce risk under a school-controlled procedure.
• Use teacher-managed sources and the school radiation rules. Record count duration and repeat background measurements. Do not extrapolate a half-life from one nucleus or from uncorrected readings.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: measured count rate falls from 100 to 55 counts per minute; background is 10. Source rates are 90 and 45, so one half-life has elapsed. Without subtraction, 55/100 would obscure the exact half in this example.

Example:

An initial source count is 160 per minute. Find it after three half-lives. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• An irradiated object is not automatically radioactive. A half-life does not predict the exact decay time of one nucleus.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Half-life predicts the exact decay time of an individual nucleus. This claim is false: An irradiated object is not automatically radioactive. A half-life does not predict the exact decay time of one nucleus.

Isotopes share proton number but differ in neutron count. A nucleus weighs less than its parts: the mass defect $\Delta m$ is the missing mass, stored as binding energy through

(masses may come in kg, MeV/c² or atomic mass units, $1\ u = 931.5\ \mathrm{MeV/c^2}$). The binding energy per nucleon curve climbs steeply to a peak near $A = 56$ (iron) and falls gently afterward: fusing light nuclei or fissioning heavy ones both release energy — the curve's interpretation is required knowledge.

What holds nuclei together against proton repulsion is the strong nuclear force: attractive, felt only between nucleons, and short-range — it fades beyond a couple of femtometres. Radioactive decay is random and spontaneous: no external trigger, no way to say which nucleus goes next.

Each decay mode changes the nucleus its own way — write the equations conserving $A$ and $Z$:

• $\alpha$: $^{A}_{Z}X \to\ ^{A-4}_{Z-2}Y + ^{4}_{2}\alpha$ (helium nucleus)
• $\beta^-$: $^{A}_{Z}X \to\ ^{A}_{Z+1}Y + ^{0}_{-1}e + \bar{\nu}$ (neutron $\to$ proton, electron, antineutrino)
• $\beta^+$: $^{A}_{Z}X \to\ ^{A}_{Z-1}Y + ^{0}_{+1}e + \nu$ (proton $\to$ neutron, positron, neutrino)
• $\gamma$: no change to $A$ or $Z$ — a photon carries away excess energy.

The three radiations differ in penetration and ionizing power: alpha (paper stops it; most ionizing), beta (a few mm of aluminium), gamma (centimetres of lead; least ionizing). The activity $A$ (becquerels) counts decays per second; a detector reads a count rate from which background radiation must be subtracted. After each half-life half the remaining nuclei decay, so after $n$ half-lives $N = N_0/2^n$.

Worked check (decay equations). Radium: $^{226}_{88}Ra \to ^{222}_{86}Rn + ^{4}_{2}\alpha$. Carbon: $^{14}_{6}C \to ^{14}_{7}N + ^{0}_{-1}e + \bar{\nu}$ — both rows conserve $A$ and $Z$ exactly.

Worked check (binding energy). Using atomic masses consistently, two hydrogen-1 atoms and two neutrons total $4.03298\ u$; a helium-4 atom weighs $4.00260\ u$. The electrons cancel, so the nuclear mass defect is $0.03038\ u$ and binding energy is $0.03038\times931.5=28.3\ \mathrm{MeV}$, or $7.07\ \mathrm{MeV}$ per nucleon.

Worked check (integer half-lives). A sample reads 5200 counts/min against a background of 200. Source alone: 5000. After three half-lives the source contributes $5000/8 = 625$, and the detector reads $625 + 200 = 825$ counts/min.

Key:

Radioactivity: mass defect binds the nucleus ($E = \Delta mc^2$, peak binding per nucleon at iron); the short-range strong force beats proton repulsion; decay is random and spontaneous with $\alpha$, $\beta^\pm$ and $\gamma$ equations conserving $A$ and $Z$; penetration falls and ionizing power rises in the order $\gamma$, $\beta$, $\alpha$; activity halves each half-life once background is removed.

Stability tracks the neutron-to-proton ratio: light stable nuclides sit near $N = Z$, heavy ones need $N > Z$ — extra neutrons dilute the proton repulsion. Above the band, $\beta^-$ pulls the nucleus back (neutron to proton); below it, $\beta^+$ does the reverse. The binding energy per nucleon is roughly constant above $A \approx 60$ — each nucleon binds only to its nearest neighbours, the saturation that is itself evidence for the strong force's short range (along with the sheer existence of nuclei against Coulomb repulsion).

Spectra carry the deeper story: alpha and gamma radiations come in discrete lines — evidence of discrete nuclear energy levels. Beta's spectrum, by contrast, is continuous: the electron takes a range of energies, which conservation laws said was impossible — unless a neutral, nearly massless partner carries the rest. The neutrino was predicted on paper years before it was found.

Quantitatively, decay follows the decay constant $\lambda$ — the probability per unit time in the limit of small $\lambda t$:

Worked check (half-life from the constant). $\lambda = 0.138\ \mathrm{day^{-1}}$: $T_{1/2} = 0.693/0.138 = 5.0$ days.

Worked check (arbitrary time). $\lambda = 0.20\ \mathrm{h^{-1}}$, $t = 5.0\ \mathrm{h}$: $N/N_0 = e^{-1.0} = 0.37$ — 37% remains, no whole number of half-lives needed. With $N_0 = 2.0\times10^{10}$ nuclei, the activity is $A = (0.20/3600)N_0 = 1.1\times10^6\ \mathrm{Bq}$ at the start.

Key:

Decay law (AHL): stability wants $N \approx Z$ rising to $N > Z$; saturated binding above $A = 60$ betrays the short-range strong force; discrete $\alpha$/$\gamma$ lines mean nuclear levels while beta's continuous spectrum demanded the neutrino; $N = N_0e^{-\lambda t}$ with $A = \lambda N$ and $T_{1/2} = \ln 2/\lambda$ handles any time interval.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A neutron can trigger a heavy nucleus to split, releasing energy and further neutrons. Whether the process becomes a chain reaction 链式反应 depends on what happens to those neutrons.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• In nuclear fission a heavy nucleus splits into lighter nuclei. Proton number and nucleon number balance in a nuclear equation. The total rest mass can decrease; the difference corresponds to released energy through E = Δmc².
• fission: Splitting of a heavy nucleus into lighter nuclei; chain reaction: A sequence in which products initiate further events.

Choose evidence that can test it

• Separate conservation of nucleon number from conservation of total energy. Released neutrons may initiate further fissions, escape or be absorbed. A controlled reactor and an uncontrolled chain reaction have different neutron-management conditions.
• Balance a supplied nuclear equation, use the specified mass data and identify the system. Analyse models or published reactor data; this is not a school attempt to produce fission or handle reactor materials.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: U-235 absorbs one neutron, forming Ba-141, Kr-92 and three neutrons. Nucleon balance: 235+1 = 141+92+3 = 236. Proton balance: 92 = 56+36. A mass decrease of 2.0×10⁻²⁸ kg corresponds to 1.8×10⁻¹¹ J using c=3.0×10⁸ m/s.

Example:

A U-235 nucleus plus one neutron produces fragments with mass numbers 141 and 92. How many neutrons balance the equation? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Mass number is not a precise mass in kilograms. Balanced integer labels alone do not calculate the released energy. Radioactive decay and neutron-induced fission are not interchangeable descriptions.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A balanced nucleon-number equation alone gives the energy release in joules. This claim is false: Mass number is not a precise mass in kilograms. Balanced integer labels alone do not calculate the released energy. Radioactive decay and neutron-induced fission are not interchangeable descriptions.

Fission splits a heavy nucleus into two medium fragments plus neutrons, releasing energy — spontaneously for a few nuclides, or induced when a neutron strikes a nucleus such as uranium-235:

Conservation holds exactly: $A$ at $236 = 141 + 92 + 3$, $Z$ at $92 = 56 + 36$. The energy is binding energy set free — the fragments sit higher on the binding-energy-per-nucleon curve than the uranium did (about $8.5$ against $7.6\ \mathrm{MeV}$ per nucleon), so roughly $235 \times 0.9 \approx 210\ \mathrm{MeV}$ per fission; direct mass bookkeeping gives the same answer.

The emitted neutrons are the engine of the chain reaction: each fission can trigger more, sustained when on average at least one product neutron causes another fission. A reactor tames it:

• Moderator (water) slows fast neutrons to thermal speeds — the speeds at which U-235 fissions readily.
• Control rods (boron, cadmium) absorb neutrons, raised or lowered to hold the reaction critical, neither growing nor dying.
• Heat exchanger carries energy from the core to water, making steam for the turbines.
• Shielding (steel and thick concrete) contains radiation from the core.

The fission products are neutron-rich and radioactive, with half-lives from seconds to millennia. They are cooled in water pools for years, then sealed — the long-term answer being deep geological repositories, whose safety over tens of thousands of years is the central consideration of nuclear waste management.

Worked check (energy per fission and per kilogram). A mass difference of $\Delta m = 0.19\ u$ per fission gives $E = 0.19 \times 931.5 = 177\ \mathrm{MeV} \approx 2.8\times10^{-11}\ \mathrm{J}$. One kilogram of U-235 holds $(1000/235) \times 6.0\times10^{23} = 2.6\times10^{24}$ nuclei, so complete fission releases about $2.6\times10^{24} \times 2.8\times10^{-11} = 7\times10^{13}\ \mathrm{J}$ in the idealized complete-fission calculation.

Key:

Fission: neutron-induced splits conserve $A$ and $Z$ and release binding energy (~200 MeV per event) because medium fragments bind tighter than uranium; emitted neutrons chain the reaction, tamed by moderator, control rods, heat exchanger and shielding; the neutron-rich products are intensely radioactive and demand cooled storage and long-term geological isolation.

Supported HL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Two stars can have the same apparent brightness while having different luminosities. Distance changes the flux 通量 reaching an observer.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Luminosity 光度 is total power emitted; flux is power received per area. A stellar spectrum carries information about surface temperature and composition. Fusion transfers energy as light nuclei combine.
• luminosity: Total emitted power; flux: Power received per unit area.

Choose evidence that can test it

• For isotropic emission without absorption, flux follows an inverse-square relationship with distance. Observed brightness alone therefore cannot establish luminosity.
• Keep distance units consistent, identify which quantities are intrinsic to the star, and distinguish observational evidence from a model of stellar evolution. Do not confuse a red giant stage with every possible final remnant.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: distance doubles while luminosity remains fixed. Use F proportional to 1/d². Flux ratio = 1/2² = 1/4. A flux of 12 units becomes 3 units. This assumes no change in absorption or source output.

Example:

A source gives flux 18 units. Find flux at three times the distance. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The Sun is not expected to become a supernova. Redshift can support cosmological expansion; it does not mean every nearby object must move away from every observer.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Apparent brightness alone always determines a star luminosity. This claim is false: The Sun is not expected to become a supernova. Redshift can support cosmological expansion; it does not mean every nearby object must move away from every observer.

A star is a balance: outward radiation pressure from the fusion-driven core against inward gravitational compression. When fusion falters, gravity wins and the star evolves.

Fusion — light nuclei merging — needs both extreme temperature (nuclei must be thrown at each other fast enough to overcome their electric repulsion) and extreme density (collisions must be frequent enough to sustain the rate). In stellar cores hydrogen fuses to helium; the mass difference is the starlight:

Stellar mass decides the story: a Sun-like star swells to a red giant, sheds its outer layers and settles as a white dwarf; a much heavier star burns fast, swells to a supergiant, and detonates as a supernova, leaving a neutron star or black hole. The elements themselves are fusion's ash — up to iron in cores, heavier atoms scattered by supernovae.

The Hertzsprung–Russell diagram maps luminosity (vertical) against surface temperature (horizontal, hot on the left). Its main regions: the main sequence running diagonally from hot-bright to cool-dim; red giants and supergiants at the top right (bright but cool, so enormous); white dwarfs at the bottom left (hot but dim, so tiny); the instability strip between. Lines of constant radius run diagonally — radius fixes the L–T combination through

so a star's radius follows from its luminosity and temperature. Temperature and composition are read from the stellar spectrum (its colour and lines).

Distances by parallax: as Earth orbits, nearby stars shift against the background by angle $p$:

with $1\ \mathrm{pc} = 3.26\ \mathrm{ly} = 2.1\times10^5\ \mathrm{AU}$.

Worked check (parallax). A star shifts by $0.25$ arcsec: $d = 1/0.25 = 4.0$ parsecs $= 4.0 \times 3.26 = 13$ light years.

Worked check (a stellar radius). A star has 100 times the Sun's luminosity at the same surface temperature: $L \propto R^2T^4$, so $R = \sqrt{100} = 10$ solar radii — a giant.

Key:

Stars and fusion: radiation pressure balances gravity while fusion lasts; fusion needs extreme temperature (Coulomb barrier) and density, turning 0.0287 u per helium nucleus into 26.7 MeV of light; mass sets evolution — red giant to white dwarf, or supergiant to supernova and neutron star or black hole; the HR diagram's main sequence, giants, supergiants, white dwarfs and instability strip map stars by luminosity and temperature with constant-radius lines from $L = 4\pi R^2\sigma T^4$; parallax measures distance as $d(\mathrm{pc}) = 1/p(\mathrm{arcsec})$.

지원되는 HL 초점. 첫 번째 평가 2025; Physics 가이드 전체 획득 (84 PDF 페이지). 남은 가이드, 평가 및 실습 요건은 기록된 보류 상태로 유지된다.

사전 요구사항: 명시된 양과 단위를 읽고, 산술 및 아래 모델 조건을 사용하십시오. 각 수업은 독립적 이전 전에 자체 방법을 개발합니다.

이는 실제 측정값이나 완성된 평가된 학습자 조사가 아닌, 원작 또는 명시적으로 허구적인 교육 예시입니다.

이 관측을 설명할 수 있는 것은 무엇인가?

• 모든 데이터 포인트 근처를 지나는 직선은 유용하지만, 그 기울기(gradiant)는 여전히 불확실할 수 있습니다. 그래프는 측정 범위 내에서 모델에 대한 증거가 됩니다.
• 예측을 먼저 제시하십시오. 비교하고자 하는 양이나 특성을 명시한 뒤, 두 설명을 구별할 수 있는 증거가 무엇인지 결정하십시오.

모델을 구축하시오

• 우연적 변동(random variation)은 반복 측정이 서로 다르게 나타나게 합니다. 계통적 오차(systematic error) 는 결과를 일관되게 편향시킵니다. 절대 불확실성(uncertainty) 은 측정값의 단위를 가지며, 상대적 또는 백분율 불확실성은 불확실성을 측정값과 비교하여 표현합니다.
• 불확실성: 측정 결과에 대한 정량적 제한; 계통오차: 일관적인 측정 편향.

이를 검증할 수 있는 증거를 선택하시오

• 곱이나 나눗셈의 경우, 최대 불확도를 근사하기 위해 분수 불확도를 더하는 것이 일반적이다. 차이의 경우 절대 불확도를 더한다. 영이 아닌 절편은 오프셋이나 불완전한 모델을 나타낼 수 있다.
• 축(axis)에 단위를 표시하고 적절한 눈금(scale)을 선택하십시오. 타당한 경우에 불확실성 막대(error bars)를 그리되, 모든 점을 연결하기보다 최선 적합 선(best-fit line)을 그리고 과정这个方法에서 요구할 때 가장 가파르고 가장 완만한 가능한 기울기를 추정하십시오.

알려진 양으로부터 시작하시오

• 알려진 값과 그 단위를 명시하십시오. 이 사례에 가정이 부합하는 식을 선택한 후 대입하기前面으로 변형하십시오.
• 주어진 값: 길이 = 50.0 mm, 불확도 1.0 mm이다. 백분율 불확도 = 절대 불확도/값 ×100 = 1.0/50.0×100 = 2.0%이다. 서로 독립적으로 측정된 두 양의 최대 백분율 불확도가 각각 2%와 3%인 경우, 합계된 최대 추정치는 5%이다.

예시:

40 cm 판독값의 절대 불확실성은 1 cm입니다. 백분율 불확실성을 구하십시오. 같은 순서를 따르십시오:已知 quantity → 모델 → 관계식 → 대입 → 단위 및 해석.

결론과 그 한계를 확인하시오

• 반복 측정값은 평균의 무작위 불확도를 줄이지만 영점 오차를 자동으로 제거하지는 않습니다. 측정 가능한 정밀도보다 더 많은 소수점 자릿수를 표기하지 마십시오.
• 원래 관찰로 돌아가십시오. 결과가 어떤 것을 지지하며, 어떤 조건을 전제로 하며, 경쟁 설명을 검증하는 방법 하나를 제시하십시오.

경고:

측정을 반복하면 항상 보정 오프셋이 제거됩니다. 이 주장은 틀렸습니다: 반복 측정값은 평균의 무작위 불확도를 줄이지만 영점 오차를 자동으로 제거하지는 않습니다. 측정 가능한 정밀도보다 더 많은 소수점 자릿수를 표기하지 마십시오.

핵심:

불확도, 기울기 및 모델 검증: 곱이나 몫의 경우, 분율 불확도를 더하는 것이 일반적인 최대 불확도 근사법입니다. 차이의 경우 절대 불확도를 더하십시오. 영이 아닌 절편은 오프셋 또는 불완전한 모델을 나타낼 수 있습니다.