Modern Physics

Around 1900, three experiments broke classical physics: the colours of glowing hot objects, the sharp lines in the light from gases, and electrons knocked out of metal by light. Each one needed a new idea: energy comes in packets. The same ideas now run solar panels, LEDs, medical scanners and nuclear power stations.

At tiny scales, energy comes in discrete packets called quanta 量子. Light is carried by photons 光子, each with energy set by its frequency:

where $h$ is Planck's constant. Wave–particle duality 波粒二象性: light and matter each show both wave behavior (interference, diffraction) and particle behavior (photons, electrons as localized hits). A particle also has a matter wavelength $\lambda=\dfrac{h}{p}$, and particles genuinely produce interference - a variation of Young's double-slit experiment 双缝实验 done with electrons builds up the same fringe pattern as light.

What a photon is. A photon has no mass and no charge, and its energy is proportional to its frequency. It travels in a straight line until it interacts with matter. In empty space every photon travels at $c=3.00\times10^{8}\ \text{m/s}$; in a material its speed is $c/n$, inversely proportional to the index of refraction. A photon also carries momentum, $p=\dfrac{h}{\lambda}$, even though it has no mass.

Matter waves. Every moving particle has a de Broglie wavelength 德布罗意波长

A smaller momentum means a longer wavelength. Quantum theory is needed when this wavelength is about as large as the system: an electron in an atom, yes; a thrown ball, no. In any bound system, such as an electron held in an atom, energy and momentum can take only certain quantized 量子化 values.

Worked example. An electron ($m=9.11\times10^{-31}\ \text{kg}$) has a wavelength of $5.0\ \text{nm}$. Its speed is

A $0.15\ \text{kg}$ ball at $40\ \text{m/s}$ has $\lambda=\dfrac{6.63\times10^{-34}}{0.15\times40}=1.1\times10^{-34}\ \text{m}$, far too small ever to show wave effects.

Worked example. Find the energy of a photon of orange light with frequency $5.0\times10^{14}\ \text{Hz}$ ($h=6.63\times10^{-34}\ \text{J s}$): $E=hf=6.63\times10^{-34}\times5.0\times10^{14}=3.3\times10^{-19}\ \text{J}$, which is about $2.1\ \text{eV}$ (dividing by $1.6\times10^{-19}$). Visible-light photons carry a few electronvolts – just right to trigger the chemistry of vision and photosynthesis.

Inside the atom. An atom has a tiny, positive nucleus made of protons and neutrons (together, nucleons 核子), with negative electrons around it. Nearly all of its mass is in the nucleus. The number of protons $Z$ decides the element. The total number of nucleons $A$ decides the isotope 同位素: carbon-12 and carbon-14 both have $Z=6$, but 6 and 8 neutrons. Nuclear notation writes this as $^{A}_{Z}\text{X}$, for example $^{14}_{\ 6}\text{C}$. An atom that has gained or lost electrons has a net charge and is called an ion 离子.

The Bohr model 玻尔模型 pictures electrons orbiting the nucleus only in certain allowed energy levels 能级. An electron can jump between levels only by absorbing or emitting a photon whose energy exactly matches the gap:

Because the levels are discrete, only specific photon energies are allowed.

Where the levels come from. Bohr treated the electron as moving in a circle, with the electric force from the nucleus as the centripetal force:

Adding the electric potential energy $U=-\dfrac{ke^2}{r}$ gives the total energy of the atom: The energy is negative because the electron is bound. A larger orbit makes $E$ less negative, so the atom's energy increases: when an atom absorbs a photon, the electron moves to a larger orbit. As $r$ grows, $K$ falls and $U$ rises (toward zero) by twice as much. Only certain orbits are allowed because the electron behaves as a standing wave: the circumference must fit a whole number of de Broglie wavelengths, $2\pi r=n\lambda$.

• An emission spectrum 发射光谱 is the set of bright lines given off when electrons drop to lower levels – each line a specific wavelength.
• An absorption spectrum 吸收光谱 is the set of dark lines where those same wavelengths are absorbed from a continuous source.

The line pattern is a fingerprint of the element, since every element has its own energy levels. An emission spectrum tells you which elements are in a light source, such as a star; an absorption spectrum tells you which elements in a gas absorbed the light.

Ground state and ionization. The lowest level is the ground state 基态; higher levels are excited states 激发态. The energy needed to remove the electron completely is the ionization energy 电离能 (the binding energy of the electron): for hydrogen in its ground state, $13.6\ \text{eV}$. It is largest from the ground state, because that level is farthest below zero.

Worked example (red light from hydrogen). A lamp contains hydrogen atoms in the ground state. To give off red light ($656\ \text{nm}$, $1.89\ \text{eV}$), an atom must first gain energy, from a collision or by absorbing a photon, and jump from $n=1$ to $n=3$ ($-13.6\to-1.51\ \text{eV}$, a gain of $12.1\ \text{eV}$). It then drops from $n=3$ to $n=2$ and emits the red photon, $-1.51-(-3.40)=1.89\ \text{eV}$, before falling back to $n=1$.

Worked example (could a photon add an electron?). A hydrogen atom absorbs a photon of $3.2\times10^{15}\ \text{Hz}$: $E=hf=6.63\times10^{-34}\times3.2\times10^{15}=2.1\times10^{-18}\ \text{J}$ ($13\ \text{eV}$). Making an electron needs its rest energy $mc^2=9.11\times10^{-31}\times(3.0\times10^{8})^2=8.2\times10^{-14}\ \text{J}$, about $40\,000$ times more. The photon cannot create an electron; it can only raise the atom's energy level.

A blackbody 黑体 emits a continuous spectrum that depends only on its temperature. Hotter objects glow brighter and peak at shorter wavelengths (red-hot to white-hot to blue-hot). Explaining this spectrum required quantized energy – a founding problem of quantum theory.

Every object turns some of its internal thermal energy into electromagnetic radiation. A blackbody is the ideal case: it absorbs all radiation that falls on it, and at a steady temperature it must emit energy at the same rate. Classical physics predicted far too much ultraviolet light; Planck fixed this by assuming energy is emitted in packets of $hf$. Two rules describe the spectrum:

• Wien's law 维恩定律: the peak wavelength falls as temperature rises, $\lambda_{\max}T=2.90\times10^{-3}\ \text{m K}$.
• Stefan–Boltzmann law 斯特藩-玻尔兹曼定律: the power emitted is proportional to the surface area and to $T^4$, $P=\sigma AT^4$.

Worked example. The Sun's surface is about $5800\ \text{K}$: $\lambda_{\max}=\dfrac{2.90\times10^{-3}}{5800}=5.0\times10^{-7}\ \text{m}=500\ \text{nm}$, in the middle of the visible range. A star with twice the temperature and the same size emits $2^4=16$ times the power, peaking at $250\ \text{nm}$ (ultraviolet).

Shining light on a metal can eject electrons – the photoelectric effect 光电效应. Key facts (which only the photon picture explains): electrons come out only if the photon's frequency exceeds a threshold, no matter how bright a dimmer, lower-frequency light is. Energy conservation gives

where $\phi$ is the metal's work function 逸出功 (the energy to free an electron). The lowest frequency that frees electrons is the threshold frequency 截止频率 $f_0=\phi/h$.

Why it proves photons exist. A wave model says brighter light should, given time, free electrons at any frequency. Experiments show the opposite. Below $f_0$, no electrons come out however bright the light. Above $f_0$, electrons come out at once even in dim light. Their maximum kinetic energy depends only on the frequency, never on the brightness. Each electron absorbs one photon, so only the energy of one photon matters.

Measuring it. In the usual set-up, light falls on a metal plate in a vacuum tube, and a variable voltage pushes back on the electrons. The voltage that just stops the fastest electrons is the stopping potential 遏止电压 $\Delta V_s$, and $K_{\max}=e\,\Delta V_s$.

Reading the graph. $K_{\max}=hf-\phi$ is a straight line. Its gradient is $h$, the same for every metal. It crosses the frequency axis at $f_0$. Extended back, it crosses the $K_{\max}$ axis at $-\phi$. Below $f_0$ the measured $K_{\max}$ is zero (no electrons), so the real data lie flat along the axis. A metal with a larger work function gives a parallel line shifted to the right.

Worked example. Light of frequency $8.0\times10^{14}\ \text{Hz}$ falls on a metal with work function $\phi=3.0\times10^{-19}\ \text{J}$. The most energetic electrons come off with

Below the threshold frequency $\phi/h$, $K_{\max}$ would be negative – meaning no electrons escape at all, however bright the light.

Worked example (the electroscope). A negatively charged electroscope has its leaves spread apart. Ultraviolet light on its metal sphere makes the leaves fall: each UV photon has more energy than the work function, so it frees electrons, the electroscope loses negative charge, and the leaves repel less. Green light does nothing, because each green photon has less energy than $\phi$. Making the green light brighter gives more photons, each with the same small energy, so still no electron escapes and the leaves still do not move.

In Compton scattering 康普顿散射, a photon collides with an electron like two particles, transferring some energy and momentum. The scattered photon comes out with less energy (longer wavelength). This is direct evidence that photons carry momentum and behave as particles.

To explain it, treat the photon as a particle with momentum $p=h/\lambda$ and apply conservation of energy and conservation of momentum to the collision. The electron recoils and gains kinetic energy, so the photon loses energy: its frequency falls, its wavelength grows and its momentum falls. The larger the angle the photon is scattered through, the more energy it gives up, so the larger the increase in wavelength. A head-on bounce straight back gives the largest change. A wave model cannot explain why the change depends on angle but not on brightness.

The nucleus 原子核 stores enormous energy. Fission 裂变 splits a heavy nucleus into lighter ones, releasing energy (nuclear reactors, bombs). Fusion 聚变 joins light nuclei into a heavier one (the Sun's power). In energy-releasing fission or fusion, the final products have slightly less total rest mass; the difference supplies energy according to $E=mc^2$. Not every possible fission or fusion reaction releases energy.

Nucleons are held together by the strong force 强相互作用, which is attractive and much stronger than the electric repulsion between protons, but acts only over nuclear distances. Every nuclear reaction conserves nucleon number, charge, energy (including rest energy, $E=mc^2$) and momentum. The released energy appears as kinetic energy of the products and as gamma photons. Some heavy nuclei split on their own (spontaneous fission); others, such as uranium-235, split only after absorbing a neutron.

Worked example. Even a tiny mass converts to a huge energy. If a nuclear reaction loses $1.0\times10^{-3}\ \text{kg}$ of mass, it releases $E=mc^2=1.0\times10^{-3}\times(3.0\times10^{8})^2=9.0\times10^{13}\ \text{J}$ – roughly the energy of $20\,000$ tonnes of TNT.

Worked example (annihilation). An electron moving with kinetic energy $K$ meets a positron at rest, and both are destroyed, producing photons. By conservation of energy, the photons share

For a slow electron $K$ is tiny next to $2mc^2$. Two photons, not one, are needed when the total momentum is close to zero, so that momentum is conserved.

Unstable nuclei undergo radioactive decay 放射性衰变, emitting:

• Alpha decay 阿尔法衰变 ($\alpha$): a helium nucleus – mass number drops by 4.
• Beta decay 贝塔衰变 ($\beta$) comes in two kinds. In beta-minus ($\beta^-$) a neutron becomes a proton, emitting an electron and an antineutrino 反中微子: $n\rightarrow p+e^-+\bar{\nu}$. In beta-plus ($\beta^+$), inside a nucleus with enough available decay energy, a proton becomes a neutron, emitting a positron 正电子 and a neutrino 中微子: $p\rightarrow n+e^++\nu$.
• Gamma decay 伽马衰变 ($\gamma$): a high-energy photon – the nucleus sheds excess energy.

The neutrino and antineutrino (symbols $\nu$, $\bar{\nu}$) are tiny, chargeless, almost-massless particles that carry off energy and keep beta decay balanced. Decay is random, but a sample halves every half-life 半衰期: after each half-life, half of the remaining nuclei have decayed. Every decay equation conserves three quantities: nucleon number (mass number), charge, and lepton number 轻子数 – an electron or neutrino counts as $+1$ lepton, a positron or antineutrino as $-1$, and the totals must match on both sides.

Decay particles. An alpha particle is a helium-4 nucleus: two protons and two neutrons ($\alpha$ or $\text{He}^{2+}$). A positron has the electron's mass but a positive charge ($e^+$ or $\beta^+$). Neutrinos and antineutrinos have no charge and almost no mass; they feel only the weak force and gravity, so they pass through matter almost without interacting. Gamma decay follows an alpha or beta decay: the new nucleus is left excited, and it drops to a lower energy state by emitting a photon. Which decay a nucleus undergoes depends on its isotope: a nucleus with too many neutrons tends toward $\beta^-$, one with too few toward $\beta^+$.

Worked example (balancing). Carbon-14 decays by $\beta^-$:

Nucleons: $14=14+0$. Charge: $6=7-1$. Leptons: $0=(+1)+(-1)$.

The decay constant. No one can predict when one nucleus will decay; decay is random, but large numbers follow a rule. The decay constant 衰变常数 $\lambda$ is the probability per second that a nucleus decays, and

Half-lives range from fractions of a second to billions of years.

Worked example. A radioactive sample has a half-life of $8.0$ days. What fraction is left after $24$ days? That is $24/8.0=3$ half-lives, so the fraction remaining is $\left(\tfrac12\right)^3=\tfrac18$ – about $12.5\%$. In an alpha decay of uranium-238 ($^{238}_{\ 92}\text{U}$), the daughter has mass number $238-4=234$ and atomic number $92-2=90$: thorium-234.

Worked example (carbon dating). Carbon-14 has $t_{1/2}=5730\ \text{y}$. A bone has $25\%$ of the carbon-14 of living bone. That is $\left(\tfrac12\right)^2$, two half-lives, so it is $2\times5730=11\,460\ \text{y}$ old. With the decay constant, for any fraction: $\lambda=\dfrac{0.693}{5730}=1.21\times10^{-4}\ \text{y}^{-1}$ and $t=\dfrac{\ln(N_0/N)}{\lambda}=\dfrac{\ln4}{1.21\times10^{-4}}=11\,500\ \text{y}$.

Answer each before reading the answers below. Take $h=6.63\times10^{-34}\ \text{J s}$, $c=3.0\times10^{8}\ \text{m/s}$ and $1\ \text{eV}=1.6\times10^{-19}\ \text{J}$.

1. Find the energy of a photon of blue light of wavelength $450\ \text{nm}$, in joules and in electronvolts.

2. That light falls on a metal with work function $2.3\ \text{eV}$. Find the maximum kinetic energy of the ejected electrons.

3. In hydrogen the $n=3$ level is at $-1.51\ \text{eV}$ and the $n=2$ level at $-3.40\ \text{eV}$. Find the energy and the wavelength of the photon emitted when an electron drops from $n=3$ to $n=2$.

4. The activity of a sample falls from $1600\ \text{Bq}$ to $100\ \text{Bq}$ in $20$ hours. Find its half-life.

5. A student says: "Brighter light ejects photoelectrons with more kinetic energy." Correct the student.

6. A proton and an electron move at the same speed. Which has the longer de Broglie wavelength? Explain.

7. A metal has a threshold frequency of $6.0\times10^{14}\ \text{Hz}$. Find its work function in eV, and the stopping potential when light of $9.0\times10^{14}\ \text{Hz}$ falls on it.

8. Complete the decay: $^{22}_{11}\text{Na}\rightarrow\,^{22}_{10}\text{Ne}+\ ?\ +\ ?$

9. A sample has a half-life of $3.0\ \text{h}$. Find its decay constant, and the fraction left after $5.0\ \text{h}$.

1. $E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{450\times10^{-9}}=4.4\times10^{-19}\ \text{J}$, which is $\dfrac{4.42\times10^{-19}}{1.6\times10^{-19}}=2.8\ \text{eV}$.

2. $K_{\max}=hf-\phi=2.76-2.30=0.46\ \text{eV}$ (about $7.4\times10^{-20}\ \text{J}$).

3. $E=-1.51-(-3.40)=1.89\ \text{eV}=3.0\times10^{-19}\ \text{J}$. Then $\lambda=\dfrac{hc}{E}=\dfrac{1.99\times10^{-25}}{3.02\times10^{-19}}=6.6\times10^{-7}\ \text{m}$, about $660\ \text{nm}$: the red line of hydrogen.

4. $1600\to800\to400\to200\to100$ is 4 half-lives in $20\ \text{h}$, so the half-life is $5.0\ \text{h}$.

5. Brighter light at the same frequency means more photons per second, so more electrons come off each second, but each photon still carries $hf$. The maximum kinetic energy $hf-\phi$ depends only on the frequency. Raise the frequency, not the brightness, to get faster electrons.

6. The electron. At the same speed it has a much smaller mass, so a smaller momentum $mv$, and $\lambda=h/(mv)$ is longer.

7. $\phi=hf_0=4.14\times10^{-15}\times6.0\times10^{14}=2.5\ \text{eV}$. $K_{\max}=4.14\times10^{-15}\times9.0\times10^{14}-2.5=3.7-2.5=1.2\ \text{eV}$, so $\Delta V_s=1.2\ \text{V}$.

8. $Z$ falls by 1 with $A$ unchanged, so a proton became a neutron: $\beta^+$ decay. The missing particles are a positron $^{0}_{+1}e$ and a neutrino $\nu$ (lepton number $0=-1+1$).

9. $\lambda=\dfrac{0.693}{3.0}=0.23\ \text{h}^{-1}$. $\dfrac{N}{N_0}=e^{-0.231\times5.0}=e^{-1.16}=0.31$.

• Photon energy is $E=hf$; below the threshold frequency no electrons are emitted however bright the light (the photoelectric effect).
• Use $K_{\max}=hf-\phi$ for the fastest photoelectrons ($\phi$ = work function).
• Electrons jump between discrete energy levels by absorbing/emitting a photon whose energy equals the level gap — the source of line spectra.
• In fission and fusion the small mass lost becomes energy via $E=mc^2$.
• Balance decay equations by conserving mass number, charge, and lepton number (write the antineutrino in $\beta^-$, the neutrino in $\beta^+$), and halve the sample every half-life ($(\tfrac12)^n$ after $n$ half-lives).

Common mistakes

• Thinking brighter light gives faster photoelectrons. Intensity changes the number of photons; each photon's energy, and so the maximum kinetic energy, depends only on frequency: $K_{\max} = hf - \phi$.
• Expecting electrons below the threshold frequency if the light is bright enough. Below $f_0 = \phi/h$ no electrons are emitted, however intense the light.
• Mixing joules and electronvolts in $E = hf$. Use $h = 6.63\times10^{-34}\ \text{J s}$ for joules or $h = 4.14\times10^{-15}\ \text{eV s}$ for electronvolts; $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.
• Saying an atom can absorb any photon. It absorbs only photons whose energy matches the gap between two of its energy levels.
• Saying that after two half-lives nothing is left. One quarter of the original nuclei remains; after $n$ half-lives, $(\tfrac12)^n$ remains.
• Leaving out the mass in $\Delta E = \Delta m\,c^2$ units: $\Delta m$ in kilograms and $c = 3.0\times10^8\ \text{m/s}$ give joules.
• Saying a larger Bohr orbit has less energy. $E=-\dfrac{ke^2}{2r}$ is negative, so a larger $r$ means a higher (less negative) energy.
• Drawing the second metal's $K_{\max}$ line steeper. Every metal has the same gradient $h$; only the intercepts change.
• Saying Compton scattering changes the photon's speed. The photon still travels at $c$; its energy, momentum and frequency fall, and its wavelength grows.