- understand that electromagnetic radiation has a particulate nature
- understand that a photon is a quantum of electromagnetic energy
- recall and use $E = hf$
- use the electronvolt (eV) as a unit of energy
- understand that a photon has momentum and that the momentum is given by $p = E/c$
Quantum physics
A-Level Physics · Topic 22
22.1
Photons: the particle nature of light
Syllabus
Source: Cambridge International syllabus
Electromagnetic radiation behaves like particles as well as like a wave. The particles of EM radiation are photons 光子 — small packets ("quanta" 量子) of EM energy that travel at the speed of light.
A photon is a packet of energy, E = hf
Energy of a photon
A photon of frequency 频率 $f$ has energy
$$E = h f,$$
where $h = 6.63 \times 10^{-34}\ \text{J s}$ is the Planck constant 普朗克常量. Using $c = f\lambda$:
$$E = \frac{h c}{\lambda}.$$
Worked example. Find the energy of a photon of green light of wavelength $500\ \text{nm}$. ($h = 6.63 \times 10^{-34}\ \text{J s}$, $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$.)
$$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{500 \times 10^{-9}} \approx 4.0 \times 10^{-19}\ \text{J}\ (\approx 2.5\ \text{eV}).$$
Higher-frequency (shorter-wavelength 波长) photons carry more energy: one $\gamma$-ray photon carries far more than one radio photon.
The electronvolt
The electronvolt 电子伏特 (eV) is a handy energy unit on the atomic scale:
$$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}.$$
It is the kinetic energy 动能 an electron 电子 gains moving through a potential difference 电势差 of 1 V. For example, a visible photon ($\lambda \approx 500\ \text{nm}$) has energy $\approx 2.5\ \text{eV}$. To go eV → J multiply by $1.60 \times 10^{-19}$; J → eV divide.
Momentum of a photon
A photon also carries momentum 动量:
$$p = \frac{E}{c} = \frac{h}{\lambda}.$$
It has zero rest mass but a non-zero momentum $E/c$. Radiation pressure (photons pushing on a surface) follows from this.
Explore
Energy of a photon
E = h·f
Photon energy is proportional to frequency — the gradient is Planck's constant h.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| photon | 光子 | guāng zi |
| quanta | 量子 | liàng zǐ |
| frequency | 频率 | pín lǜ |
| Planck constant | 普朗克常量 | pǔ lǎng kè cháng liàng |
| wavelength | 波长 | bō cháng |
| electronvolt | 电子伏特 | diàn zi fú tè |
| kinetic energy | 动能 | dòng néng |
| electron | 电子 | diàn zi |
| potential difference | 电势差 | diàn shì chà |
| momentum | 动量 | dòng liàng |
22.2
Photoelectric effect
Syllabus
- understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation
- understand and use the terms threshold frequency and threshold wavelength
- explain photoelectric emission in terms of photon energy and work function energy
- recall and use $hf = \Phi + \frac{1}{2}m{v_{\text{max}}}^2$
- explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity
Source: Cambridge International syllabus
Solar cells use the photoelectric effect to turn light into electricity.
When EM radiation of high enough frequency hits a metal, electrons are emitted. These are photoelectrons 光电子, and the effect is the photoelectric effect 光电效应.
One photon gives its energy $hf$ to one electron: part frees it (the work function $\Phi$), the rest is the electron's KE
A charged zinc plate loses its charge — the gold leaf falls — when ultraviolet light shines on it
Threshold frequency and work function
Each metal has a lowest photon frequency, the threshold frequency 极限频率 $f_{0}$, below which no electrons come out, however bright the light. The work function 逸出功 $\Phi$ is the least energy needed to free an electron from the surface:
$$\Phi = h f_{0}.$$
Different metals have different work functions (about $2$–$5\ \text{eV}$).
Einstein's photoelectric equation
One photon gives all its energy to one electron. If the photon energy $hf$ is more than the work function, the electron escapes with kinetic energy up to a maximum:
$$h f = \Phi + \tfrac{1}{2} m v_{\text{max}}^{2}, \qquad\text{so}\qquad \tfrac{1}{2} m v_{\text{max}}^{2} = h(f - f_{0}).$$
Worked example. A metal has a work function of $2.0\ \text{eV}$. Light made of photons of energy $3.5\ \text{eV}$ shines on it. Find the maximum kinetic energy of the photoelectrons.
$$\tfrac{1}{2}mv_{\text{max}}^{2} = hf - \Phi = 3.5 - 2.0 = 1.5\ \text{eV}\ (= 2.4 \times 10^{-19}\ \text{J}).$$
So the maximum KE of photoelectrons depends linearly on frequency, not on brightness.
The maximum kinetic energy of photoelectrons rises linearly with frequency, reaching zero at the threshold frequency $f_0$
Why the wave model fails
A wave model predicts that brightness should set the electrons' kinetic energy, and that emission should happen at any frequency given enough time. But experiments show:
- no emission below the threshold frequency, however bright.
- immediate emission at or above the threshold, even when dim.
- maximum KE depends on frequency, not brightness.
- the number of photoelectrons (the current) depends on brightness.
The photon model explains this: light arrives as photons each of energy $hf$. One photon–electron interaction either has enough energy to free the electron ($hf \geq \Phi$) or it does not.
Why max KE is fixed but current grows with brightness
A brighter beam of the same frequency has more photons per second, but each still carries $hf$. So the maximum KE of any electron is $hf - \Phi$ (set by $f$ only), while the rate of emission (the current) grows with the number of photons, i.e. with brightness. Doubling the brightness doubles the current but does not change the maximum KE.
Explore
The photoelectric effect
KEmax = h·f − φ
Max KE is a straight line in frequency, with intercept −φ (the work function).
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| photoelectron | 光电子 | guāng diàn zi |
| photoelectric effect | 光电效应 | guāng diàn xiào yìng |
| threshold frequency | 极限频率 | jí xiàn pín lǜ |
| work function | 逸出功 | yì chū gōng |
22.3
Wave–particle duality
Syllabus
- understand that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature
- describe and interpret qualitatively the evidence provided by electron diffraction for the wave nature of particles
- understand the de Broglie wavelength as the wavelength associated with a moving particle
- recall and use $\lambda = h/p$
Source: Cambridge International syllabus
The photoelectric effect is strong evidence for the particle nature of light. But interference 干涉 (Young's double slit, the diffraction grating 衍射光栅) and diffraction 衍射 show its wave nature. So light has both wave and particle sides — this is wave–particle duality 波粒二象性.
De Broglie hypothesis
If a wave can act like particles, perhaps particles can act like waves. De Broglie proposed that any moving particle has a de Broglie wavelength 德布罗意波长:
$$\lambda = \frac{h}{p},$$
where $p = mv$. Example: an electron at $v = 4.9 \times 10^{7}\ \text{m s}^{-1}$ has $p = 4.46 \times 10^{-23}\ \text{kg m s}^{-1}$, so $\lambda = 1.49 \times 10^{-11}\ \text{m} \approx 0.015\ \text{nm}$ — close to atomic spacings.
Electron diffraction
When electrons are fired at a crystal lattice 晶格 (e.g. thin graphite), they make a diffraction pattern of bright rings on a screen — exactly what waves of wavelength $\lambda = h/p$ would do. This is direct evidence for the wave nature of particles (electron diffraction 电子衍射): only waves diffract, yet electrons do.
A faster electron has more momentum, so a shorter de Broglie wavelength, which diffracts less — the rings move closer together. Slowing the electrons spreads the rings apart. To calculate: $p = \sqrt{2 m E_{\text{K}}}$, and for an electron accelerated through p.d. $V$, $E_{\text{K}} = eV$, so $\lambda = h/\sqrt{2m_{e} e V}$.
Electrons fired at graphite form a ring diffraction pattern — only waves diffract, so electrons behave as waves
Worked example. An electron is accelerated from rest through a p.d. of $2500\ \text{V}$. Find its de Broglie wavelength. ($m_{e} = 9.11 \times 10^{-31}\ \text{kg}$, $e = 1.6 \times 10^{-19}\ \text{C}$, $h = 6.63 \times 10^{-34}\ \text{J s}$.)
Its kinetic energy is $E_{\text{K}} = eV$, so $\lambda = \dfrac{h}{\sqrt{2 m_{e} e V}}$:
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2(9.11 \times 10^{-31})(1.6 \times 10^{-19})(2500)}} \approx 2.5 \times 10^{-11}\ \text{m}.$$
This is close to the spacing between atoms in a crystal, which is why the electrons diffract off the graphite.
Explore
The de Broglie wavelength
λ = h/p
A particle's wavelength is inversely proportional to its momentum — faster, heavier particles have shorter waves.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| interference | 干涉 | gān shè |
| diffraction grating | 衍射光栅 | yǎn shè guāng shān |
| diffraction | 衍射 | yǎn shè |
| wave–particle duality | 波粒二象性 | bō lì èr xiàng xìng |
| de Broglie wavelength | 德布罗意波长 | dé bù luó yì bō cháng |
| crystal lattice | 晶格 | jīng gé |
| electron diffraction | 电子衍射 | diàn zi yǎn shè |
22.4
Energy levels in atoms
Syllabus
- understand that there are discrete electron energy levels in isolated atoms (e.g. atomic hydrogen)
- understand the appearance and formation of emission and absorption line spectra
- recall and use $hf = E_1 - E_2$
Source: Cambridge International syllabus
In an isolated atom, electrons can only sit at certain discrete 分立 energy levels 能级 — never in between. The lowest is the ground state 基态; the others are excited states 激发态.
By convention, energies are written negative, with $E = 0$ for an electron just free of the atom. For hydrogen the ground state is $E_{1} = -13.6\ \text{eV}$; higher states approach zero.
The electron energy levels of hydrogen are discrete and negative, with the ground state at $-13.6\ \text{eV}$
Emission spectrum
When an electron drops from a higher level $E_{2}$ to a lower level $E_{1}$, it emits one photon of energy
$$h f = E_{2} - E_{1}.$$
(Both energies are negative; their difference is positive.) Because the levels are discrete, only certain photon energies — and so certain wavelengths — come out. The emission spectrum 发射光谱 is a set of sharp bright lines on a dark background, one line per transition 跃迁. The pattern is a "fingerprint" of the element.
The emission spectrum of hydrogen is a set of sharp bright lines on a dark background
Real emission spectra of the elements: each one is a unique set of bright lines -- a fingerprint of that element
Absorption spectrum
When white light passes through a cool gas, photons whose energy exactly matches an upward transition are absorbed. The light then shows dark lines on a bright background — the absorption spectrum 吸收光谱. The dark lines sit at the same wavelengths as the emission lines of the same gas.
Dark absorption lines in the Sun's spectrum mark the wavelengths absorbed by cooler gas
Calculations
For a transition between two known levels:
$$hf = E_{2} - E_{1}, \qquad \lambda = \frac{hc}{E_{2} - E_{1}}.$$
Work in consistent units — convert eV to joules (× $1.60 \times 10^{-19}$) before finding $\lambda$ in metres, or use $hc \approx 1240\ \text{eV nm}$ for a quick estimate.
Explore
Make an element's spectral lines
An electron dropping between fixed energy levels emits a photon of exactly the gap's energy — a fixed wavelength and colour. Each jump is one line of the element's barcode.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| discrete | 分立 | fēn lì |
| energy level | 能级 | néng jí |
| ground state | 基态 | jī tài |
| excited state | 激发态 | jī fā tài |
| emission spectrum | 发射光谱 | fā shè guāng pǔ |
| transition | 跃迁 | yuè qiān |
| absorption spectrum | 吸收光谱 | xī shōu guāng pǔ |
22.4
Exam tips
- Photon energy $E = hf = hc/\lambda$; the photoelectric equation is $hf = \Phi + \frac{1}{2}mv_{max}^2$.
- The photoelectric effect shows light is quantised: below the threshold frequency, no electrons are emitted whatever the intensity.
- de Broglie $\lambda = h/p$ gives particles a wavelength; electron diffraction is the evidence.
- Energy levels are discrete; an emitted photon's energy equals the difference between two levels.