Nuclear physics · ⁨פיזיקה גרעינית⁩

Einstein's special relativity gives the famous link (mass-energy equivalence 质能等价):

where $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$. A mass $m$ matches an energy 能量 $E$ — the two can change into each other. For a mass change $\Delta m$:

Worked example. The star Sirius loses mass through nuclear fusion at $1.09 \times 10^{11}\ \text{kg s}^{-1}$. Find the power it radiates.

Every kilogram that disappears leaves as energy: $P = c^{2} \times (\text{mass lost per second}) = (3.00 \times 10^{8})^{2}(1.09 \times 10^{11}) = 9.8 \times 10^{27}\ \text{W}$. This is the star's luminosity 光度 (Topic 25); the Sun's is $3.8 \times 10^{26}\ \text{W}$, so it loses about $4$ million tonnes a second. The same idea in reverse: a power station producing $1\ \text{GW}$ for a year converts $E/c^{2} = 3.2 \times 10^{16}/9.0 \times 10^{16} = 0.35\ \text{kg}$ of mass, which is why the fuel weighs almost the same afterwards.

In nuclear physics the masses are tiny but $c^{2}$ is huge, so a small mass change means a large energy. A mass change of $1\ \text{u}$ ($1.661 \times 10^{-27}\ \text{kg}$) matches $\Delta E \approx 1.49 \times 10^{-10}\ \text{J}$. This gives a conversion you will use again and again:

A nuclear reaction 核反应 is written like

with nucleon number 核子数 conserved (top numbers: $14 + 4 = 17 + 1$) and charge conserved (bottom numbers: $7 + 2 = 8 + 1$) — this is conservation of charge 电荷守恒. Use these to fill in an unknown: identify the species, then balance the top and bottom numbers.

The decay equations. Alpha decay removes $^{4}_{2}\text{He}$, so $A$ falls by 4 and $Z$ by 2: $^{211}_{84}\text{Po} \to {}^{207}_{82}\text{Pb} + {}^{4}_{2}\text{He}$. Beta-minus decay turns a neutron into a proton, emitting an electron and an antineutrino: $^{15}_{6}\text{C} \to {}^{15}_{7}\text{N} + {}^{0}_{-1}\text{e} + \bar{\nu}$ ($A$ unchanged, $Z$ up by 1). Beta-plus decay turns a proton into a neutron, emitting a positron 正电子 and a neutrino 中微子: $^{18}_{9}\text{F} \to {}^{18}_{8}\text{O} + {}^{0}_{+1}\text{e} + \nu$ ($Z$ down by 1). Gamma emission changes neither number. In a chain of decays just keep the books: a nucleus W that emits $\beta^{-}$, then $\alpha$, then $\beta^{-}$ ends with $A - 4$ and $Z + 1 - 2 + 1 = Z$, an isotope of W. Check both lines of every equation you complete, and remember the neutrino: the mark scheme includes it.

The mass of a nucleus 原子核 is less than the total mass of its separate protons 质子 and neutrons 中子. The difference is the mass defect 质量亏损 $\Delta m$:

By $E = mc^{2}$, this "missing" mass was released as energy when the nucleus formed. To pull the nucleus fully apart you must put that energy back — the binding energy 结合能 $B$:

Worked example. A helium-4 nucleus has a mass defect of $\Delta m = 0.0304\ \text{u}$. Find its binding energy. ($1\ \text{u}$ corresponds to $931\ \text{MeV}$.)

A more tightly bound nucleus has a larger mass defect and larger binding energy. The binding energy per nucleon 比结合能 is $B/A$ (usually in MeV per nucleon) — a measure of how tightly each nucleon is held, useful for comparing nuclides.

The two-mark definitions. The mass defect of a nucleus is the difference between the total mass of its separate nucleons and the mass of the nucleus. The binding energy is the minimum energy required to separate the nucleus into its individual nucleons (equivalently, the energy released when the nucleus is formed from separate nucleons). Say "separate nucleons" or "individual protons and neutrons"; "the energy holding the nucleus together" scores nothing. Binding energy is released, not stored: the nucleus has less energy than its parts.

Worked example. The masses are: proton $1.007\,276\ \text{u}$, neutron $1.008\,665\ \text{u}$, polonium-212 nucleus $211.945\,4\ \text{u}$. Find the mass defect and the binding energy per nucleon of $^{212}_{84}\text{Po}$.

$Z = 84$ protons and $N = 212 - 84 = 128$ neutrons: total $84 \times 1.007\,276 + 128 \times 1.008\,665 = 84.611\,2 + 129.109\,1 = 213.720\,3\ \text{u}$. Mass defect $\Delta m = 213.720\,3 - 211.945\,4 = 1.774\,9\ \text{u}$. Binding energy $= 1.774\,9 \times 931.5 = 1653\ \text{MeV}$, so per nucleon $1653/212 = 7.80\ \text{MeV}$, on the falling part of the curve. Keep every decimal place until the subtraction: the defect is a small difference of two large numbers.

Binding energy per nucleon vs nucleon number

A graph of $B/A$ against $A$ has a typical shape:

• for light nuclei ($A < 20$), $B/A$ rises quickly (with a spike at the very stable $^{4}_{2}\text{He}$).
• around $A \sim 56$ (iron), $B/A$ reaches its maximum of about $8.8\ \text{MeV}$. Iron-56 is the most stable nucleus.
• for heavy nuclei ($A > 100$), $B/A$ falls slowly, to about $7.5\ \text{MeV}$ for uranium.

So the curve is dome-shaped, rising to iron then falling.

Sketching the curve. The exam gives blank axes ($A$ from 1 to 250, $B/A$ up to about $9\ \text{MeV}$) and marks: a steep rise from near zero at $A = 1$, a maximum near $A = 56$ at about $8.8\ \text{MeV}$, then a slow, gentle fall to about $7.5\ \text{MeV}$ at $A = 238$. Do not start the curve at the origin exactly (hydrogen-1 has no binding energy but is a single point), do not make the fall as steep as the rise, and do not let the curve reach zero on the right. Asked to mark a nucleus that undergoes alpha decay, put X on the far right, $A > 200$; a nucleus that undergoes fusion goes at the far left, $A < 10$; both are at low $B/A$, moving up the curve when they react.

Energy is released when nuclei move towards the iron peak — by joining light nuclei or splitting heavy ones.

Nuclear fusion

Nuclear fusion 核聚变 joins two light nuclei into one heavier nucleus:

The product has greater binding energy per nucleon than the reactants, so energy is released. Fusion powers stars. It needs very high temperatures (millions of kelvin) so the nuclei have enough kinetic energy 动能 to beat their electrostatic 静电 repulsion and get close enough for the strong nuclear force 强核力 to take over.

Nuclear fission

Nuclear fission 核裂变 splits a heavy nucleus into two lighter ones:

The products have higher binding energy per nucleon than $^{235}$U, so energy is released. The extra neutrons can cause more fissions — a chain reaction 链式反应 in a large enough mass of fuel (the critical mass 临界质量). This is the basis of nuclear power and weapons.

"Describe the differences between fission and fusion." Fission: a heavy nucleus (large $A$) splits into two lighter nuclei of roughly similar mass, usually after absorbing a neutron, releasing further neutrons. Fusion: two light nuclei (small $A$) join to form one heavier nucleus; it needs very high temperature and pressure to overcome the electrostatic repulsion between the nuclei. Both release energy, but per kilogram of fuel fusion releases more.

"Explain, with reference to the curve, why energy is released." In both processes the products lie higher on the binding-energy-per-nucleon curve than the reactants: each nucleon ends up more tightly bound, so the total binding energy increases. The extra binding energy is released (as the kinetic energy of the products and as photons), and the total mass of the products is less than that of the reactants by $\Delta E/c^{2}$. A nucleus near the peak (iron) can release energy by neither process, which is why the stars' fusion stops at iron.

Calculating the energy released

1. find the total mass of the reactants.
2. find the total mass of the products.
3. mass change $\Delta m = m_{\text{reactants}} - m_{\text{products}}$ (positive when energy is released).
4. energy released $\Delta E = c^{2} \Delta m$.

In kg this gives joules; in atomic mass units use $\Delta E\ (\text{MeV}) = \Delta m\ (\text{u}) \times 931$.

Worked example. In a nuclear reaction the total mass decreases by $0.020\ \text{u}$. Find the energy released.

Worked example (fusion). The mass defect of deuterium $^{2}_{1}\text{H}$ is $0.002\,388\ \text{u}$ and that of helium-4 is $0.030\,377\ \text{u}$. Find the energy released when two deuterium nuclei fuse to form one helium-4 nucleus.

Energy released $=$ (binding energy of the products) $-$ (binding energy of the reactants) $= [0.030\,377 - 2 \times 0.002\,388] \times 931.5 = 0.025\,601 \times 931.5 = 23.8\ \text{MeV}$ ($3.82 \times 10^{-12}\ \text{J}$). Mass defects can be used directly like this because the number of nucleons is the same on both sides; the difference in the defects is the mass converted. Per kilogram of deuterium this is $5.7 \times 10^{14}\ \text{J}$, about a million times a chemical fuel.

Worked example (fission). $^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n}$. Masses: U-235 $235.043\,9\ \text{u}$, n $1.008\,665\ \text{u}$, Ba-141 $140.914\,4\ \text{u}$, Kr-92 $91.926\,2\ \text{u}$. Find the energy released.

Reactants: $235.043\,9 + 1.008\,665 = 236.052\,6\ \text{u}$. Products: $140.914\,4 + 91.926\,2 + 3 \times 1.008\,665 = 235.866\,6\ \text{u}$. $\Delta m = 0.186\,0\ \text{u}$, so $\Delta E = 0.186\,0 \times 931.5 = 173\ \text{MeV} = 2.8 \times 10^{-11}\ \text{J}$ per fission. Count the three neutrons on the right and the one on the left: forgetting one changes the answer by a whole nucleon mass.

Worked example (alpha decay and momentum). A stationary $^{238}_{92}\text{U}$ nucleus decays to $^{234}_{90}\text{Th}$ by emitting an $\alpha$-particle; the total kinetic energy released is $4.27\ \text{MeV}$. Find the kinetic energy of the $\alpha$-particle.

Momentum is conserved and the parent was at rest, so the $\alpha$ and the thorium nucleus have equal and opposite momenta $p$. With $E_{\text{K}} = p^{2}/2m$, the energies are in the inverse ratio of the masses: $E_{\alpha}/E_{\text{Th}} = m_{\text{Th}}/m_{\alpha} = 234/4$. So $E_{\alpha} = 4.27 \times 234/238 = 4.20\ \text{MeV}$ and the thorium recoil 反冲 takes only $0.07\ \text{MeV}$. The $\alpha$-particles from a given decay are all emitted with this one energy, which is the sign that the energy is shared between just two bodies.

Worked example (a radioactive power source). A space probe is powered by $0.874\ \text{kg}$ of plutonium-238, half-life $87.7$ years, each decay releasing $5.59\ \text{MeV}$. Find the power available at launch.

Number of nuclei: $N = 0.874/(238 \times 1.661 \times 10^{-27}) = 2.21 \times 10^{24}$. Decay constant: $\lambda = 0.693/(87.7 \times 3.156 \times 10^{7}) = 2.50 \times 10^{-10}\ \text{s}^{-1}$. Activity: $A = \lambda N = 5.53 \times 10^{14}\ \text{Bq}$. Power $= A \times E = 5.53 \times 10^{14} \times 5.59 \times 1.60 \times 10^{-13} = 490\ \text{W}$, falling to half after $87.7$ years. A nuclide with a shorter half-life would give more power per kilogram but would not last the mission; polonium-210 ($138$ days) would be far more powerful at first and useless within two years.

Random and spontaneous

Radioactive decay is:

• spontaneous 自发 — it happens with no outside trigger, and the rate is not changed by temperature, pressure or chemical state; and
• random 随机 — you cannot predict when a given nucleus will decay, only the probability that it decays in a time.

Evidence for randomness: the count rate fluctuates. A Geiger counter 盖革计数器 next to a source clicks at uneven intervals — never a steady stream — although the long-run mean rate is well-defined.

The two-mark definitions. Radioactive decay is the spontaneous and random emission of a particle ($\alpha$ or $\beta$) or a photon ($\gamma$) from an unstable nucleus. Spontaneous means the decay is not affected by external factors: temperature, pressure, chemical state, or the presence of other nuclei. Random means it is impossible to predict which nucleus will decay next, or when a given nucleus will decay; only a probability can be given. Evidence for randomness: the count rate fluctuates from one interval to the next, even for a source whose activity is not changing over the experiment.

Activity and decay constant

For $N$ undecayed nuclei of a radionuclide 放射性核素, the rate of decay is

• $A$ is the activity 活度 — decays per unit time. Unit: becquerel 贝克勒尔 (Bq) $= \text{s}^{-1}$.
• $\lambda$ is the decay constant 衰变常数 — the probability per unit time that a nucleus decays. Unit: $\text{s}^{-1}$.

$\lambda$ is fixed for a nuclide; a larger sample (larger $N$) has proportionally larger activity.

The one-mark definitions. Activity is the number of decays (of nuclei) per unit time, or the rate of decay. The decay constant is the probability per unit time that a (given) nucleus will decay. Not "the rate of decay": that is the activity. Note the units are the same, $\text{s}^{-1}$, but $A$ counts events and $\lambda$ is a probability per second.

Worked example. Fluorine-18 has a half-life of $110$ minutes. Show that its decay constant is $1.05 \times 10^{-4}\ \text{s}^{-1}$, and find the activity of $2.1 \times 10^{-12}\ \text{kg}$ of fluorine-18.

$\lambda = 0.693/(110 \times 60) = 1.05 \times 10^{-4}\ \text{s}^{-1}$. The number of nuclei is the mass divided by the mass of one nucleus: $N = 2.1 \times 10^{-12}/(18 \times 1.661 \times 10^{-27}) = 7.0 \times 10^{13}$. So $A = \lambda N = 1.05 \times 10^{-4} \times 7.0 \times 10^{13} = 7.4 \times 10^{9}\ \text{Bq}$. A tiny mass gives a huge activity because the half-life is short; the same mass of uranium-238 (half-life $4.5 \times 10^{9}$ years) would give about $10^{-5}\ \text{Bq}$.

Exponential decay

Since $\lambda$ is the fractional decay rate, $\dfrac{dN}{dt} = -\lambda N$, whose solution is an exponential decay 指数衰减:

Because $A = \lambda N$, the activity (and any count rate 计数率 proportional to it) decays the same way:

Why exponential? For each nucleus, $\lambda$ is a fixed probability per unit time, independent of the others and of the nucleus's age. So the same fraction decays in each time interval, which gives exponential decay.

Writing the explanation (three marks). (1) The decay constant is the probability per unit time of decay and is the same for every nucleus of the nuclide, whatever its age. (2) So the rate of decay, the activity, is proportional to the number of undecayed nuclei present: $A = \lambda N$. (3) A rate of change proportional to the quantity itself gives an exponential change; equivalently, the same fraction of the remaining nuclei decays in every equal time interval, so the number never reaches zero but halves in every half-life.

Half-life

The half-life 半衰期 $t_{1/2}$ is the time for the number of undecayed nuclei (or the activity, or the count rate) to fall to half. From $N = N_{0} e^{-\lambda t}$ with $N = N_{0}/2$:

Worked example. A radioactive isotope has a half-life of $6.0$ hours. Find its decay constant.

Convert the half-life to seconds: $6.0\ \text{h} = 21\,600\ \text{s}$. Then

A larger decay constant means a shorter half-life. After $n$ half-lives the surviving fraction is $(1/2)^{n}$; after 5 half-lives only about 3% remains.

The definition of half-life. The time taken for the number of undecayed nuclei (or the activity) of a sample to fall to half its initial value. "Half the atoms decay" is accepted; "half the sample disappears" is not, since the decayed nuclei are still there as the daughter product.

Reading a two-isotope graph. When X decays to a stable Y, $N_{\text{Y}} = N_{0} - N_{\text{X}} = N_{0}(1 - e^{-\lambda t})$. The half-life is the time at which the curves cross ($N_{\text{X}} = N_{\text{Y}} = N_{0}/2$), or the time for $N_{\text{X}}$ to halve. If the graph gives the initial number $N_{0}$ and the initial mass $m$, the nucleon number follows from $m = N_{0} A u$: for $m = 7.3 \times 10^{-4}\ \text{kg}$ and $N_{0} = 2.0 \times 10^{21}$, $A = m/(N_{0} u) = 7.3 \times 10^{-4}/(2.0 \times 10^{21} \times 1.661 \times 10^{-27}) = 220$.

Worked example. A sample of a single radioactive isotope has an activity of $180\ \text{Bq}$ at $t = 0$ and $45\ \text{Bq}$ at $t = 8.4$ minutes. Find the half-life and the decay constant, and the activity after a further $8.4$ minutes.

$45/180 = 1/4 = (1/2)^{2}$: two half-lives in $8.4$ minutes, so $t_{1/2} = 4.2\ \text{min} = 252\ \text{s}$ and $\lambda = 0.693/252 = 2.8 \times 10^{-3}\ \text{s}^{-1}$. After another two half-lives the activity is $45/4 = 11\ \text{Bq}$. When the ratio is not a neat power of two, use $t = \ln(A_{0}/A)/\lambda$: for the activity to fall from $180$ to $50\ \text{Bq}$ takes $\ln(3.6)/2.75 \times 10^{-3} = 466\ \text{s}$. Always convert the half-life to seconds before finding $\lambda$ if the activity is in becquerels.

Finding $\lambda$ from data

Given $A_{0}$ and $A$ at time $t$:

Taking logs of $A = A_{0} e^{-\lambda t}$ gives $\ln A = \ln A_{0} - \lambda t$, so a plot of $\ln A$ against $t$ is a straight line with gradient $-\lambda$. Use this with several data points.

Count rate is not activity. A detector records only the radiation that reaches it and is absorbed in it: a fraction set by the solid angle it covers, by absorption in the air and in the source itself, and by its efficiency. So the measured count rate is smaller than the activity, but proportional to it, and the half-life obtained from a count-rate graph is correct. Subtract the background radiation 本底辐射 (measured with the source removed) from every reading before taking ratios or logarithms; an unsubtracted background makes the curve flatten and the half-life appear too long.

Tracers in medicine. Fluorine-18 and oxygen-15 are $\beta^{+}$ emitters used as tracers 示踪剂 in PET scanning (Topic 24): the positron annihilates with an electron, giving two gamma photons that leave in opposite directions and reveal where the tracer is. A short half-life ($110$ minutes, $2$ minutes) is chosen so that the activity falls quickly after the scan and the dose to the patient stays small, at the price of having to make the isotope close to the hospital and use it at once.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• $E = mc^{2}$ with $1\ \text{u} = 931.5\ \text{MeV}$: work in u and MeV for reactions, then convert to joules only if asked ($1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$).
• Energy released $=$ (total mass before $-$ total mass after) $\times c^{2}$, or (binding energy after $-$ binding energy before). Mass defects can be subtracted directly when the nucleon count is unchanged.
• The curve: steep rise, peak $8.8\ \text{MeV}$ near $A = 56$, gentle fall. Products higher on the curve means energy released; fusion on the left, fission (and α-decay) on the right.
• Balance every equation twice: nucleon numbers along the top, proton numbers along the bottom; β decays carry a neutrino or antineutrino.
• $A = \lambda N$, $\lambda = 0.693/t_{1/2}$ with $t_{1/2}$ in seconds; $N$ from mass: $N = m/(A u)$. Ratios of $1/2$, $1/4$, $1/8$ mean whole half-lives; otherwise $t = \ln(x_{0}/x)/\lambda$.
• Explanations are marked on the words: constant probability, rate proportional to number, same fraction per interval; fluctuating count rate for randomness; unaffected by external conditions for spontaneous.

Common mistakes

• Defining binding energy as "the energy holding the nucleus together" or "the energy stored in the nucleus"; it is the energy to separate the nucleons.
• Subtracting masses the wrong way round and reporting a negative energy release, or forgetting the neutron(s) on one side of a fission equation.
• Drawing the binding-energy curve falling as steeply as it rises, or reaching zero at large $A$.
• Using a half-life in minutes or years with an activity in becquerels; convert to seconds first.
• Confusing the decay constant (a probability per second) with the activity (decays per second).
• Saying "half the sample disappears" for a half-life, or that after two half-lives nothing is left.
• Giving "random" as "it happens at any time" without "cannot predict which nucleus or when", or "spontaneous" without "unaffected by external factors".
• Treating the count rate as the activity, or forgetting to subtract background.
• Assuming the α-particle and the recoil nucleus share the energy equally; they share the momentum equally.