Quantum and Nuclear Physics

The Photoelectric Effect

Observations

When light of sufficiently high frequency is incident on a metal surface, electrons are ejected. Key experimental observations:

1. Threshold frequency: Electrons are emitted only if $f \ge f_0$, regardless of intensity.
2. Instantaneous emission: No detectable time delay between illumination and emission.
3. Maximum kinetic energy of photoelectrons depends on frequency, not intensity.
4. More intensity (at $f \ge f_0$) produces more photoelectrons, not faster ones.

These observations cannot be explained by the classical wave model of light.

Einstein's Explanation (1905)

Light consists of discrete packets of energy called photons. Each photon has energy:

$E = hf$

where $h = 6.626 \times 10^{-34}\,\mathrm{J\,s}$ is Planck's constant and $f$ is the frequency.

A single photon can eject at most one electron. The photon gives its entire energy to the electron. Some energy overcomes the work function $\Phi$ (minimum energy to escape the metal); the remainder becomes kinetic energy:

$E_k = hf - \Phi$

This is Einstein's photoelectric equation.

Threshold Frequency

At the threshold, $E_k = 0$:

$hf_0 = \Phi \implies f_0 = \frac{\Phi}{h}$

Stopping Potential

The stopping potential $V_s$ is the minimum voltage needed to prevent the most energetic photoelectrons from reaching the collector:

$eV_s = E_{k,\max} = hf - \Phi$

Experimental Graphs

Graph	Gradient	$y$-intercept	$x$-intercept
$E_{k,\max}$ vs $f$	$h$	$-\Phi$	$f_0$
$V_s$ vs $f$	$h/e$	$-\Phi/e$	$f_0$

Example. Light of wavelength $400\,\mathrm{nm}$ is incident on a zinc surface with work function $\Phi = 4.3\,\mathrm{eV}$. Find the maximum kinetic energy of the emitted electrons.

$E_{\mathrm{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19}\,\mathrm{J} = 3.11\,\mathrm{eV}$

$E_k = 3.11 - 4.3 = -1.19\,\mathrm{eV}$

Since $E_k \lt 0$, no photoelectrons are emitted. The photon energy is below the work function.

---

The Photon Model

Photon Energy and Momentum

A photon has energy $E = hf = \dfrac{hc}{\lambda}$ and momentum:

$p = \frac{E}{c} = \frac{h}{\lambda}$

Electron Volt

The electron volt is a unit of energy convenient for atomic-scale physics:

$1\,\mathrm{eV} = 1.602 \times 10^{-19}\,\mathrm{J}$

Useful constant:

$hc = 1240\,\mathrm{eV\,nm}$

This allows quick conversion: for a photon of wavelength $500\,\mathrm{nm}$:

$E = \frac{1240}{500} = 2.48\,\mathrm{eV}$

Compton Scattering

When an X-ray photon collides with a free electron, it transfers some energy and the photon's wavelength increases. The Compton shift is:

$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$

where $\theta$ is the scattering angle and $\dfrac{h}{m_e c} = 2.43 \times 10^{-12}\,\mathrm{m}$ is the Compton wavelength of the electron. This demonstrates the particle nature of electromagnetic radiation.

Worked Example: Compton Scattering

An X-ray photon of wavelength $0.0500\,\mathrm{nm}$ is scattered at $90^\circ$ by a free electron. Find the wavelength of the scattered photon.

$\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) = (2.43 \times 10^{-12})(1 - \cos 90^\circ)$

$\Delta\lambda = (2.43 \times 10^{-12})(1 - 0) = 2.43 \times 10^{-12}\,\mathrm{m} = 0.00243\,\mathrm{nm}$

$\lambda' = 0.0500 + 0.00243 = 0.05243\,\mathrm{nm}$

---

Wave-Particle Duality

De Broglie Hypothesis (1924)

Every particle has an associated wavelength:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

where $p = mv$ is the momentum of the particle.

Example. Find the de Broglie wavelength of an electron accelerated through $100\,\mathrm{V}$.

$E_k = eV = 100\,\mathrm{eV} = 1.6 \times 10^{-17}\,\mathrm{J}$

$p = \sqrt{2m_e E_k} = \sqrt{2(9.109 \times 10^{-31})(1.6 \times 10^{-17})} = 5.40 \times 10^{-24}\,\mathrm{kg\,m/s}$

$\lambda = \frac{6.626 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10}\,\mathrm{m} = 0.123\,\mathrm{nm}$

This is comparable to atomic spacing, explaining why electron diffraction is observable.

Electron Diffraction

The Davisson-Germer experiment (1927) confirmed the wave nature of electrons. An electron beam directed at a nickel crystal produced a diffraction pattern consistent with the de Broglie wavelength. The constructive interference condition is:

$d\sin\theta = n\lambda$

This is the same equation as for X-ray diffraction (Bragg's law), but with $\lambda = h/(mv)$.

Significance

Wave-particle duality is a fundamental property of nature. All matter exhibits wave-like properties, but the effect is only significant at atomic and subatomic scales. For macroscopic objects, the de Broglie wavelength is far too small to detect.

---

Atomic Energy Levels

Bohr Model

Niels Bohr proposed that electrons in atoms occupy discrete energy levels (orbitals). An electron can transition between levels by absorbing or emitting a photon of energy exactly equal to the energy difference:

$hf = \Delta E = E_{\mathrm{upper}} - E_{\mathrm{lower}}$

Hydrogen Energy Levels

$E_n = -\frac{13.6\,\mathrm{eV}}{n^2}, \qquad n = 1, 2, 3, \ldots$

• $n = 1$: ground state ($-13.6\,\mathrm{eV}$)
• $n = 2$: first excited state ($-3.4\,\mathrm{eV}$)
• $n = \infty$: ionisation ($0\,\mathrm{eV}$)

The ionisation energy of hydrogen is $13.6\,\mathrm{eV}$.

Emission and Absorption Spectra

Emission spectrum: When excited atoms de-excite, they emit photons at discrete frequencies, producing bright lines on a dark background.

Absorption spectrum: When white light passes through cool gas, atoms absorb photons at specific frequencies, producing dark lines on a continuous spectrum.

For hydrogen, the wavelengths of the spectral lines are given by the Rydberg formula:

$\frac{1}{\lambda} = R_H\!\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

where $R_H = 1.097 \times 10^7\,\mathrm{m^{-1}}$ is the Rydberg constant, $n_i \gt n_f$.

Series	$n_f$	Region
Lyman	1	Ultraviolet
Balmer	2	Visible
Paschen	3	Infrared

Example. Find the wavelength of the first Balmer line ($n_i = 3 \to n_f = 2$).

$\frac{1}{\lambda} = 1.097 \times 10^7\!\left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{5}{36} = 1.524 \times 10^6\,\mathrm{m^{-1}}$

$\lambda = 656\,\mathrm{nm}$

This is the characteristic red line of the hydrogen spectrum ($\mathrm{H}\alpha$).

Worked Example: Hydrogen Spectral Lines

An electron in a hydrogen atom transitions from $n = 4$ to $n = 2$.

Find the energy, wavelength, and frequency of the emitted photon.

Energy of levels:

$E_4 = \frac{-13.6}{16} = -0.85\,\mathrm{eV}, \quad E_2 = \frac{-13.6}{4} = -3.40\,\mathrm{eV}$

Photon energy:

$\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55\,\mathrm{eV}$

Wavelength:

$\lambda = \frac{hc}{\Delta E} = \frac{1240}{2.55} = 486\,\mathrm{nm}$

Frequency:

$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{486 \times 10^{-9}} = 6.17 \times 10^{14}\,\mathrm{Hz}$

This is the blue-green $\mathrm{H}\beta$ line in the Balmer series.

Is the electron excited or de-excited? The electron moves from a higher energy level ($n = 4$) to a lower one ($n = 2$), so this is de-excitation and a photon is emitted.

---

Radioactive Decay

Nuclear Stability

Nuclei are stable only for certain combinations of protons ($Z$) and neutrons ($N$). For light nuclei, stability requires $N \approx Z$. For heavier nuclei, more neutrons are needed ($N \gt Z$) to counteract the increasing Coulomb repulsion between protons.

Types of Radioactive Decay

Decay	Emission	Change	Example
Alpha ($\alpha$)	$^4_2\mathrm{He}$ (helium nucleus)	$Z \to Z - 2$, $A \to A - 4$	$^{238}_{92}\mathrm{U} \to ^{234}_{90}\mathrm{Th} + \alpha$
Beta-minus ($\beta^-$)	$e^-$ (electron) + $\bar{\nu}_e$	$n \to p$: $Z \to Z + 1$, $A$ unchanged	$^{14}_6\mathrm{C} \to ^{14}_7\mathrm{N} + e^- + \bar{\nu}_e$
Beta-plus ($\beta^+$)	$e^+$ (positron) + $\nu_e$	$p \to n$: $Z \to Z - 1$, $A$ unchanged	$^{11}_6\mathrm{C} \to ^{11}_5\mathrm{B} + e^+ + \nu_e$
Gamma ($\gamma$)	High-energy photon	No change in $Z$ or $A$	Excited nucleus de-excites

The Decay Law

Radioactive decay is a random process governed by:

$N = N_0 e^{-\lambda t}$

where $N$ is the number of undecayed nuclei at time $t$, $N_0$ is the initial number, and $\lambda$ is the decay constant.

Activity (rate of decay): $A = -\dfrac{dN}{dt} = \lambda N = A_0 e^{-\lambda t}$, measured in becquerels ($\mathrm{Bq}$), where $1\,\mathrm{Bq} = 1$ decay per second.

Half-Life

The half-life $t_{1/2}$ is the time for half the nuclei to decay:

$N_0 e^{-\lambda t_{1/2}} = \frac{N_0}{2} \implies t_{1/2} = \frac{\ln 2}{\lambda}$

After $n$ half-lives: $N = N_0 \left(\dfrac{1}{2}\right)^n$.

Example. Cobalt-60 has a half-life of $5.27\,\mathrm{years}$. A sample initially has activity $800\,\mathrm{Bq}$. Find the activity after $15.81\,\mathrm{years}$.

Number of half-lives: $n = 15.81 / 5.27 = 3$.

$A = 800 \times \left(\frac{1}{2}\right)^3 = 100\,\mathrm{Bq}$

Worked Example: Decay Constant and Half-Life

A radioactive isotope has a half-life of $8.0\,\mathrm{days}$. A sample contains $4.0 \times 10^{15}$ undecayed nuclei at $t = 0$.

(a) Find the decay constant.

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.0 \times 24 \times 3600} = 1.00 \times 10^{-6}\,\mathrm{s}^{-1}$

(b) Find the initial activity.

$A_0 = \lambda N_0 = (1.00 \times 10^{-6})(4.0 \times 10^{15}) = 4.0 \times 10^9\,\mathrm{Bq}$

(c) How long until the activity falls to $1.0 \times 10^8\,\mathrm{Bq}$?

$A = A_0 e^{-\lambda t} \implies 1.0 \times 10^8 = 4.0 \times 10^9 \times e^{-(1.00 \times 10^{-6})t}$

$e^{-(1.00 \times 10^{-6})t} = 0.025 \implies -(1.00 \times 10^{-6})t = \ln(0.025)$

$t = \frac{-3.689}{1.00 \times 10^{-6}} = 3.69 \times 10^6\,\mathrm{s} \approx 42.7\,\mathrm{days}$

---

Nuclear Reactions

Nuclear Fission

A heavy nucleus splits into two (or more) lighter nuclei, releasing energy and neutrons. A chain reaction occurs when released neutrons induce further fission events.

$^{235}_{92}\mathrm{U} + ^1_0\mathrm{n} \to ^{141}_{56}\mathrm{Ba} + ^{92}_{36}\mathrm{Kr} + 3\,^1_0\mathrm{n} + \mathrm{energy}$

Critical mass: the minimum mass of fissile material needed to sustain a chain reaction.

Conditions for a controlled chain reaction:

• Fuel exceeds the critical mass.
• Neutrons are moderated (slowed) to increase the fission cross-section.
• Control rods absorb excess neutrons to regulate the reaction rate.

Nuclear Fusion

Light nuclei combine to form a heavier nucleus, releasing energy. Fusion powers stars and is the basis of the proton-proton chain:

$^1_1\mathrm{H} + ^1_1\mathrm{H} \to ^2_1\mathrm{H} + e^+ + \nu_e$

$^2_1\mathrm{H} + ^1_1\mathrm{H} \to ^3_2\mathrm{He} + \gamma$

$^3_2\mathrm{He} + ^3_2\mathrm{He} \to ^4_2\mathrm{He} + 2\,^1_1\mathrm{H}$

Net: $4\,^1_1\mathrm{H} \to ^4_2\mathrm{He} + 2e^+ + 2\nu_e + 2\gamma + 26.7\,\mathrm{MeV}$

Fusion requires extremely high temperatures ($\sim 10^7\,\mathrm{K}$) to overcome Coulomb repulsion.

---

Mass-Energy Equivalence

Einstein's Equation

$E = mc^2$

The mass defect $\Delta m$ of a nucleus is the difference between the mass of the separated nucleons and the mass of the bound nucleus:

$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$

This mass defect represents the energy released when the nucleus was formed.

Binding Energy

The binding energy of a nucleus is the energy required to completely separate it into its constituent nucleons:

$E_b = \Delta m \cdot c^2$

The binding energy per nucleon $E_b/A$ is a measure of nuclear stability. It peaks around iron-56 ($\sim 8.8\,\mathrm{MeV/nucleon}$), which is the most stable nucleus.

Nucleus	Binding Energy per Nucleon (MeV)
$^2_1\mathrm{H}$ (deuterium)	1.11
$^4_2\mathrm{He}$	7.07
$^{56}_{26}\mathrm{Fe}$	8.79
$^{235}_{92}\mathrm{U}$	7.59

Implications:

• Fission of heavy nuclei ($A \gt 56$) releases energy because the products have higher binding energy per nucleon.
• Fusion of light nuclei ($A \lt 56$) releases energy for the same reason.
• Iron-56 is the most stable nucleus; neither fission nor fusion of iron releases energy.

Calculating Binding Energy

Example. Calculate the binding energy of the helium-4 nucleus.

Given: $m_p = 1.00728\,\mathrm{u}$, $m_n = 1.00867\,\mathrm{u}$, $m_{\mathrm{He}} = 4.00260\,\mathrm{u}$, $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$.

$\Delta m = 2(1.00728) + 2(1.00867) - 4.00260 = 0.03030\,\mathrm{u}$

$E_b = 0.03030 \times 931.5 = 28.2\,\mathrm{MeV}$

$\frac{E_b}{A} = \frac{28.2}{4} = 7.07\,\mathrm{MeV/nucleon}$

Energy Released in Reactions

Example. Find the energy released in the fission reaction:

$^{235}_{92}\mathrm{U} + ^1_0\mathrm{n} \to ^{141}_{56}\mathrm{Ba} + ^{92}_{36}\mathrm{Kr} + 3\,^1_0\mathrm{n}$

Masses: $m_{\mathrm{U-235}} = 235.0439\,\mathrm{u}$, $m_{\mathrm{Ba-141}} = 140.9139\,\mathrm{u}$, $m_{\mathrm{Kr-92}} = 91.8973\,\mathrm{u}$, $m_n = 1.0087\,\mathrm{u}$.

Reactants: $235.0439 + 1.0087 = 236.0526\,\mathrm{u}$. Products: $140.9139 + 91.8973 + 3(1.0087) = 235.8373\,\mathrm{u}$.

$\Delta m = 236.0526 - 235.8373 = 0.2153\,\mathrm{u}$

$E = 0.2153 \times 931.5 \approx 200.6\,\mathrm{MeV}$

Worked Example: Binding Energy per Nucleon

Calculate the binding energy per nucleon of lithium-7 ($^7_3\mathrm{Li}$).

Given: $m_p = 1.00728\,\mathrm{u}$, $m_n = 1.00867\,\mathrm{u}$, $m_{\mathrm{Li-7}} = 7.01600\,\mathrm{u}$, $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$.

Lithium-7 has $Z = 3$ protons and $N = 4$ neutrons.

$\Delta m = 3(1.00728) + 4(1.00867) - 7.01600 = 3.02184 + 4.03468 - 7.01600 = 0.04052\,\mathrm{u}$

$E_b = 0.04052 \times 931.5 = 37.74\,\mathrm{MeV}$

$\frac{E_b}{A} = \frac{37.74}{7} = 5.39\,\mathrm{MeV/nucleon}$

This is lower than the binding energy per nucleon of helium-4 ($7.07\,\mathrm{MeV/nucleon}$), which reflects the exceptional stability of the helium nucleus (an "alpha particle" with a filled shell structure).

---

Heisenberg Uncertainty Principle

Statement

It is fundamentally impossible to simultaneously know both the position and momentum of a particle with arbitrary precision:

$\Delta x \cdot \Delta p \ge \frac{\hbar}{2}$

where $\hbar = \dfrac{h}{2\pi} = 1.055 \times 10^{-34}\,\mathrm{J\,s}$.

This is not a limitation of measurement technology but a fundamental property of nature. It arises directly from the wave nature of matter: a well-defined wavelength (precise momentum) requires an extended wave (uncertain position).

Energy-Time Uncertainty

$\Delta E \cdot \Delta t \ge \frac{\hbar}{2}$

This allows virtual particle-antiparticle pairs to briefly exist, provided $\Delta E \cdot \Delta t$ is sufficiently small.

Worked Example: Heisenberg Uncertainty Principle

An electron is confined within a region of width $\Delta x = 1.0 \times 10^{-10}\,\mathrm{m}$ (roughly the diameter of a hydrogen atom).

Find the minimum uncertainty in its momentum.

$\Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2(1.0 \times 10^{-10})} = 5.28 \times 10^{-25}\,\mathrm{kg\,m/s}$

Find the corresponding minimum uncertainty in velocity.

$\Delta v = \frac{\Delta p}{m_e} = \frac{5.28 \times 10^{-25}}{9.11 \times 10^{-31}} = 5.80 \times 10^5\,\mathrm{m/s}$

This is a significant fraction of the speed of light, showing that confining an electron to atomic dimensions implies a very large uncertainty in its velocity -- consistent with the probabilistic nature of electron behaviour in atoms.

---

Pair Production and Annihilation

Pair Production

A photon can convert into a particle-antiparticle pair (e.g. $e^- + e^+$) provided its energy exceeds the total rest energy of the pair:

$E_{\mathrm{photon}} \ge 2m_e c^2 = 1.022\,\mathrm{MeV}$

Momentum must also be conserved, which requires the presence of a nearby nucleus to absorb recoil momentum. Pair production cannot occur in empty space.

Pair Annihilation

When a particle meets its antiparticle, they annihilate, converting their combined rest mass into photon energy. For an electron-positron pair at rest:

$2m_e c^2 = 2(0.511\,\mathrm{MeV}) = 1.022\,\mathrm{MeV}$

This energy is carried by two photons (to conserve momentum), each with energy $0.511\,\mathrm{MeV}$, emitted in opposite directions.

---

Nuclear Physics Applications

Carbon Dating

Living organisms continuously exchange carbon with the environment, maintaining a constant ratio of $^{14}\mathrm{C}$ to $^{12}\mathrm{C}$. After death, $^{14}\mathrm{C}$ decays with a half-life of $5730\,\mathrm{years}$. The age of a sample is determined from the remaining $^{14}\mathrm{C}$:

$N = N_0 e^{-\lambda t} \implies t = \frac{1}{\lambda}\ln\!\left(\frac{N_0}{N}\right) = \frac{t_{1/2}}{\ln 2}\ln\!\left(\frac{N_0}{N}\right)$

Example. A sample has $25\%$ of the original $^{14}\mathrm{C}$. Find its age.

$t = \frac{5730}{0.693}\ln(4) = 5730 \times 2 = 11460\,\mathrm{years}$

Nuclear Medicine

• Technetium-99m ($t_{1/2} = 6.01\,\mathrm{h}$): gamma emitter used in diagnostic imaging.
• Iodine-131 ($t_{1/2} = 8.02\,\mathrm{d}$): beta emitter used to treat thyroid conditions.
• Cobalt-60 ($t_{1/2} = 5.27\,\mathrm{y}$): gamma emitter used in radiotherapy.

Nuclear Reactor Components

Component	Function
Fuel ($^{235}\mathrm{U}$)	Undergoes fission, releasing energy
Moderator (water, graphite)	Slows neutrons to thermal energies
Control rods (boron, cadmium)	Absorb neutrons to control reaction rate
Coolant (water, $\mathrm{CO}_2$)	Transfers heat from reactor core
Shielding (concrete, lead)	Absorbs radiation for safety

---

Wave Functions and Probability

The Schrodinger Equation

The time-independent Schrodinger equation for a particle of mass $m$ in a potential $V(x)$:

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi$

where $\psi(x)$ is the wave function and $E$ is the energy eigenvalue.

Probability Interpretation

The wave function $\psi$ has no direct physical meaning, but $|\psi(x)|^2$ gives the probability density for finding the particle at position $x$:

$P(x)\,dx = |\psi(x)|^2\,dx$

The total probability must be unity (normalisation):

$\int_{-\infty}^{\infty} |\psi(x)|^2\,dx = 1$

Particle in a Box

For a particle confined to a one-dimensional box of length $L$ ($V = 0$ inside, $V = \infty$ outside):

$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right), \qquad n = 1, 2, 3, \ldots$

$E_n = \frac{n^2 h^2}{8mL^2}$

Key features: energy is quantised, the ground state has non-zero energy ($n = 1$), and the particle has non-zero probability of being found at any position inside the box.

Quantum Tunneling

A particle with energy $E \lt V_0$ has a non-zero probability of passing through a potential barrier of height $V_0$. The transmission coefficient decreases exponentially with barrier width $w$:

$T \approx e^{-2\kappa w}$

where $\kappa = \dfrac{\sqrt{2m(V_0 - E)}}{\hbar}$.

Quantum tunneling is responsible for alpha decay, tunnel diodes, and scanning tunnelling microscopy.

---

Standard Model Overview

Classification of Particles

Fermions (half-integer spin): matter particles.

• Quarks (six flavours: up, down, charm, strange, top, bottom): experience all four forces.
• Leptons (electron, muon, tau + their neutrinos): experience gravity, EM, weak force.

Bosons (integer spin): force carriers.

• Photon ($\gamma$): electromagnetic force.
• Gluon ($g$): strong nuclear force.
• $W^+$, $W^-$, $Z^0$: weak nuclear force.
• Higgs boson: gives mass to $W$, $Z$ bosons and fermions.

The Four Fundamental Forces

Force	Mediator	Relative Strength	Range
Strong	Gluon	1	$\sim 10^{-15}\,\mathrm{m}$
Electromagnetic	Photon	$\sim 10^{-2}$	Infinite
Weak	$W$, $Z$ bosons	$\sim 10^{-6}$	$\sim 10^{-18}\,\mathrm{m}$
Gravitational	Graviton (hypothetical)	$\sim 10^{-38}$	Infinite

---

Feynman Diagrams

Purpose

Feynman diagrams are pictorial representations of particle interactions. Each diagram represents a term in a perturbation expansion of the quantum field theory amplitude.

Conventions

• Straight lines with arrows: fermions (matter particles; arrows reversed for antiparticles).
• Wavy lines: photons.
• Curly lines: gluons.
• Dashed lines: $W$ or $Z$ bosons, or Higgs.
• Vertices represent interactions; conservation laws apply at each vertex.

Example: Electron-Positron Annihilation

$e^- + e^+ \to \gamma + \gamma$: The electron and positron annihilate into a virtual photon, which produces two real photons. The diagram has two incoming fermion lines, one internal photon line, and two outgoing photon lines.

warning

Common Pitfall

The binding energy curve shows a peak at iron-56, but the curve is relatively flat around this peak. Elements from nickel to lead all have binding energies per nucleon in the range $7.5$--$8.8\,\mathrm{MeV/nucleon}$. Do not assume that fission of elements lighter than iron always absorbs energy; the actual threshold depends on the specific reaction.

---

Problem Set

Problem 1

Light of wavelength $250\,\mathrm{nm}$ is incident on a sodium surface with work function $\Phi = 2.28\,\mathrm{eV}$. Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential.

Solution

Photon energy:

$E = \frac{hc}{\lambda} = \frac{1240}{250} = 4.96\,\mathrm{eV}$

Maximum kinetic energy:

$E_{k,\max} = E - \Phi = 4.96 - 2.28 = 2.68\,\mathrm{eV}$

Stopping potential:

$eV_s = E_{k,\max} \implies V_s = 2.68\,\mathrm{V}$

If you get this wrong, revise: The Photoelectric Effect section.

Problem 2

Find the de Broglie wavelength of a neutron moving at $2.0 \times 10^4\,\mathrm{m/s}$. ($m_n = 1.675 \times 10^{-27}\,\mathrm{kg}$)

Solution

$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(1.675 \times 10^{-27})(2.0 \times 10^4)} = \frac{6.626 \times 10^{-34}}{3.35 \times 10^{-23}}$

$\lambda = 1.98 \times 10^{-11}\,\mathrm{m} = 0.0198\,\mathrm{nm}$

This is comparable to X-ray wavelengths, explaining why neutron diffraction is used to study crystal structures.

If you get this wrong, revise: Wave-Particle Duality section.

Problem 3

A hydrogen atom absorbs a photon of wavelength $97.3\,\mathrm{nm}$. Determine the transition involved (initial and final energy levels).

Solution

Photon energy:

$E = \frac{1240}{97.3} = 12.75\,\mathrm{eV}$

Energy levels: $E_n = -13.6/n^2$.

The photon is absorbed, so the electron moves to a higher level:

$\Delta E = E_{n_f} - E_{n_i} = 12.75\,\mathrm{eV}$

If the electron starts from $n = 1$ ($E_1 = -13.6\,\mathrm{eV}$):

$E_{n_f} = -13.6 + 12.75 = -0.85\,\mathrm{eV}$

$-0.85 = \frac{-13.6}{n_f^2} \implies n_f^2 = 16 \implies n_f = 4$

The transition is $n = 1 \to n = 4$ (absorption, Lyman series).

If you get this wrong, revise: Atomic Energy Levels section.

Problem 4

Strontium-90 has a half-life of $28.8\,\mathrm{years}$. A sample initially contains $2.0 \times 10^{20}$ atoms. How many atoms remain after $100\,\mathrm{years}$? What is the activity at that time?

Solution

Decay constant:

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{28.8 \times 365.25 \times 24 \times 3600} = 7.64 \times 10^{-10}\,\mathrm{s}^{-1}$

Number remaining:

$N = N_0 e^{-\lambda t}$

$t = 100 \times 365.25 \times 24 \times 3600 = 3.156 \times 10^9\,\mathrm{s}$

$\lambda t = (7.64 \times 10^{-10})(3.156 \times 10^9) = 2.411$

$N = 2.0 \times 10^{20} \times e^{-2.411} = 2.0 \times 10^{20} \times 0.0897 = 1.79 \times 10^{19}$

Activity:

$A = \lambda N = (7.64 \times 10^{-10})(1.79 \times 10^{19}) = 1.37 \times 10^{10}\,\mathrm{Bq}$

If you get this wrong, revise: Radioactive Decay section.

Problem 5

Complete the following nuclear equation and identify the type of decay:

$^{234}_{90}\mathrm{Th} \to ^{234}_{91}\mathrm{Pa} + \,?$

Solution

Conserving mass number: $234 = 234 + A \implies A = 0$

Conserving atomic number: $90 = 91 + Z \implies Z = -1$

The emitted particle has $A = 0$ and $Z = -1$, which is an electron: $e^{-}$ (or $\beta^{-}$).

This is beta-minus decay, in which a neutron converts to a proton, emitting an electron and an antineutrino:

$^{234}_{90}\mathrm{Th} \to ^{234}_{91}\mathrm{Pa} + e^{-} + \bar{\nu}_e$

If you get this wrong, revise: Types of Radioactive Decay table.

Problem 6

Calculate the binding energy of carbon-12 ($^{12}_{6}\mathrm{C}$). Given: $m_p = 1.00728\,\mathrm{u}$, $m_n = 1.00867\,\mathrm{u}$, $m_{\mathrm{C-12}} = 12.00000\,\mathrm{u}$ (by definition of the atomic mass unit), $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$.

Solution

Carbon-12 has $Z = 6$ protons and $N = 6$ neutrons.

$\Delta m = 6(1.00728) + 6(1.00867) - 12.00000 = 6.04368 + 6.05202 - 12.00000 = 0.09570\,\mathrm{u}$

$E_b = 0.09570 \times 931.5 = 89.1\,\mathrm{MeV}$

$\frac{E_b}{A} = \frac{89.1}{12} = 7.43\,\mathrm{MeV/nucleon}$

If you get this wrong, revise: Mass-Energy Equivalence section.

Problem 7

Find the minimum energy a photon must have to produce an electron-positron pair. If the photon has exactly this energy, can pair production occur? Explain.

Solution

Minimum energy:

$E_{\min} = 2m_e c^2 = 2(0.511\,\mathrm{MeV}) = 1.022\,\mathrm{MeV}$

If the photon has exactly $1.022\,\mathrm{MeV}$, pair production cannot occur in free space because momentum cannot be conserved. The photon has momentum $p = E/c$, but the electron-positron pair at rest has zero momentum. A nearby nucleus must be present to absorb the recoil momentum. The photon energy must be greater than $1.022\,\mathrm{MeV}$ for pair production to actually occur.

If you get this wrong, revise: Pair Production section.

Problem 8

An electron is confined in a one-dimensional box of length $L = 0.50\,\mathrm{nm}$. Find the energy of the ground state and the first excited state. What is the wavelength of a photon emitted when the electron transitions from $n = 2$ to $n = 1$?

Solution

Ground state ($n = 1$):

$E_1 = \frac{h^2}{8mL^2} = \frac{(6.626 \times 10^{-34})^2}{8(9.109 \times 10^{-31})(0.50 \times 10^{-9})^2}$

$E_1 = \frac{4.390 \times 10^{-67}}{1.822 \times 10^{-49}} = 2.41 \times 10^{-18}\,\mathrm{J} = 15.0\,\mathrm{eV}$

First excited state ($n = 2$):

$E_2 = 4E_1 = 60.0\,\mathrm{eV}$

Photon energy for $n = 2 \to n = 1$:

$\Delta E = 60.0 - 15.0 = 45.0\,\mathrm{eV}$

Wavelength:

$\lambda = \frac{1240}{45.0} = 27.6\,\mathrm{nm}$

This is in the ultraviolet region.

If you get this wrong, revise: Particle in a Box section.

Problem 9

A sample of wood from an archaeological site has $^{14}\mathrm{C}$ activity that is $35\%$ of the activity of a living sample. Estimate the age of the wood. ($t_{1/2}$ of $^{14}\mathrm{C} = 5730\,\mathrm{years}$)

Solution

$N = N_0 e^{-\lambda t} \implies 0.35 = e^{-\lambda t}$

$-\lambda t = \ln(0.35) = -1.050$

$t = \frac{1.050}{\lambda} = \frac{1.050 \times t_{1/2}}{\ln 2} = \frac{1.050 \times 5730}{0.693} = 8680\,\mathrm{years}$

The wood is approximately $8700\,\mathrm{years}$ old.

If you get this wrong, revise: Carbon Dating section.

Problem 10

In a Compton scattering experiment, a photon is scattered at $180^\circ$ (backscattered) by a free electron. If the incident photon has wavelength $0.0100\,\mathrm{nm}$, find the wavelength of the scattered photon and the kinetic energy transferred to the electron.

Solution

Wavelength shift:

$\Delta\lambda = \frac{h}{m_e c}(1 - \cos 180^\circ) = (2.43 \times 10^{-12})(1 - (-1)) = 4.86 \times 10^{-12}\,\mathrm{m}$

$\lambda' = 0.0100 \times 10^{-9} + 4.86 \times 10^{-12} = 1.486 \times 10^{-11}\,\mathrm{m} = 0.01486\,\mathrm{nm}$

Energy of incident photon:

$E_i = \frac{1240}{0.0100} = 124000\,\mathrm{eV} = 124\,\mathrm{keV}$

Energy of scattered photon:

$E_f = \frac{1240}{0.01486} = 83446\,\mathrm{eV} = 83.4\,\mathrm{keV}$

Kinetic energy of electron:

$E_k = E_i - E_f = 124 - 83.4 = 40.6\,\mathrm{keV}$

If you get this wrong, revise: Compton Scattering section.

Problem 11

A proton is confined within a nucleus of radius approximately $5.0 \times 10^{-15}\,\mathrm{m}$. Estimate the minimum kinetic energy of the proton using the Heisenberg uncertainty principle. ($m_p = 1.67 \times 10^{-27}\,\mathrm{kg}$)

Solution

$\Delta x \approx 5.0 \times 10^{-15}\,\mathrm{m}$

$\Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2(5.0 \times 10^{-15})} = 1.055 \times 10^{-20}\,\mathrm{kg\,m/s}$

Using $E_k \approx \frac{(\Delta p)^2}{2m}$:

$E_k \approx \frac{(1.055 \times 10^{-20})^2}{2(1.67 \times 10^{-27})} = \frac{1.113 \times 10^{-40}}{3.34 \times 10^{-27}} = 3.33 \times 10^{-14}\,\mathrm{J}$

$E_k \approx \frac{3.33 \times 10^{-14}}{1.602 \times 10^{-19}} \approx 208\,\mathrm{keV}$

This shows that confinement energy of nucleons is on the order of MeV, consistent with nuclear binding energies.

If you get this wrong, revise: Heisenberg Uncertainty Principle section.

Problem 12

A nuclear power plant produces $3.0 \times 10^9\,\mathrm{W}$ of thermal power. Each fission of $^{235}\mathrm{U}$ releases approximately $200\,\mathrm{MeV}$. Calculate the number of fissions per second and the mass of $^{235}\mathrm{U}$ consumed per day.

Solution

Energy per fission:

$E_{\mathrm{fission}} = 200\,\mathrm{MeV} = 200 \times 1.602 \times 10^{-13} = 3.20 \times 10^{-11}\,\mathrm{J}$

Fissions per second:

$\mathrm{Rate} = \frac{P}{E_{\mathrm{fission}}} = \frac{3.0 \times 10^9}{3.20 \times 10^{-11}} = 9.38 \times 10^{19}\,\mathrm{fissions/s}$

Fissions per day:

$N = 9.38 \times 10^{19} \times 86400 = 8.10 \times 10^{24}$

Mass of $^{235}\mathrm{U}$ (molar mass $\approx 235\,\mathrm{g/mol}$):

$m = \frac{N \times 235}{N_A} = \frac{8.10 \times 10^{24} \times 235}{6.022 \times 10^{23}} = \frac{1.904 \times 10^{27}}{6.022 \times 10^{23}} \approx 3160\,\mathrm{g} \approx 3.2\,\mathrm{kg}$

Approximately $3.2\,\mathrm{kg}$ of $^{235}\mathrm{U}$ is consumed per day.

If you get this wrong, revise: Nuclear Fission and Mass-Energy Equivalence sections.