Quantum and Nuclear Physics

1. Quantum Physics (Topic 12, HL)

1.1 The Photoelectric Effect

When electromagnetic radiation is incident on a metal surface, electrons may be ejected. This phenomenon is called the photoelectric effect. The experimental observations cannot be explained by classical wave theory:

• There is a threshold frequency $f_0$ below which no electrons are emitted, regardless of intensity.
• The maximum kinetic energy of emitted electrons depends on the frequency, not the intensity.
• Electrons are emitted instantaneously, even at very low intensities.

Einstein explained these observations by proposing that light consists of discrete packets of energy called photons, each with energy:

$E = hf$

where $h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$ is Planck's constant and $f$ is the frequency.

A photon can transfer all its energy to a single electron. The work function $\phi$ is the minimum energy required to liberate an electron from the metal surface. Einstein's photoelectric equation gives the maximum kinetic energy of emitted electrons:

$E_{k,\mathrm{max}} = hf - \phi$

Since $\phi = hf_0$ (the threshold frequency), this can also be written as:

$E_{k,\mathrm{max}} = h(f - f_0)$

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

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

:::info[Data Booklet Reference] Key photoelectric equations:

$E = hf$

$E_{k,\mathrm{max}} = hf - \phi$

$\phi = hf_0$

:::

1.2 Wave-Particle Duality

The de Broglie hypothesis states that all matter exhibits wave-like properties. A particle with momentum $p$ has an associated wavelength:

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

where $m$ is the relativistic mass and $v$ is the velocity. For non-relativistic particles, $\lambda = \frac{h}{m_0 v}$ where $m_0$ is the rest mass.

Evidence for wave nature of matter:

• Electron diffraction: Electrons passing through a thin crystal produce a diffraction pattern identical to that produced by X-rays. The diffraction condition is $d\sin\theta = n\lambda$, confirming that electrons have a wavelength given by the de Broglie equation.
• Davisson-Germer experiment: Electrons scattered from a nickel crystal showed intensity maxima at specific angles consistent with diffraction of waves.

Evidence for particle nature of light:

• The photoelectric effect (photons transfer discrete quanta of energy).
• Compton scattering (see Section 1.5): X-ray photons scatter off electrons, behaving as particles in collisions.

The wave-particle duality is a fundamental principle: all quantum objects exhibit both wave-like and particle-like properties depending on the type of observation performed.

1.3 Energy Levels and Spectral Lines

The Bohr model of the atom postulates that electrons orbit the nucleus only in certain allowed energy levels. When an electron transitions between energy levels, a photon is emitted or absorbed with energy equal to the difference between the two levels:

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

For hydrogen, the energy levels are given by:

$E_n = \frac{-13.6\,\mathrm{eV}}{n^2}$

where $n = 1, 2, 3, \ldots$ is the principal quantum number.

Emission spectra are produced when electrons fall from higher to lower energy levels, emitting photons of specific frequencies. Each element produces a unique line spectrum, a "fingerprint" used in spectroscopy.

Absorption spectra are produced when white light passes through a cool gas. Electrons absorb photons of specific frequencies to jump to higher energy levels, producing dark lines in a continuous spectrum.

The Rydberg formula for hydrogen gives the wavelengths of spectral lines:

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

where $R = 1.097 \times 10^7\,\mathrm{m}^{-1}$ is the Rydberg constant, $n_i$ is the initial level, and $n_f$ is the final level.

Spectral series:

Series	$n_f$	Region
Lyman	1	Ultraviolet
Balmer	2	Visible
Paschen	3	Infrared
Brackett	4	Infrared

Worked Example: Photon Emission in Hydrogen

An electron in a hydrogen atom transitions from $n = 4$ to $n = 2$. Calculate the energy, frequency, and wavelength of the emitted photon, and identify the spectral series.

Energy of photon:

$\Delta E = E_4 - E_2 = \frac{-13.6}{16} - \frac{-13.6}{4} = -0.85 - (-3.40) = 2.55\,\mathrm{eV}$

Convert to joules: $\Delta E = 2.55 \times 1.60 \times 10^{-19} = 4.08 \times 10^{-19}\,\mathrm{J}$

Frequency:

$f = \frac{\Delta E}{h} = \frac{4.08 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.15 \times 10^{14}\,\mathrm{Hz}$

Wavelength:

$\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{6.15 \times 10^{14}} = 4.88 \times 10^{-7}\,\mathrm{m} = 488\,\mathrm{nm}$

This is in the visible range (blue-green), part of the Balmer series ($n_f = 2$).

If you get this wrong, revise: Section 1.3.

1.4 Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle is a fundamental limit on the precision with which certain pairs of physical properties can be known simultaneously. For position and momentum:

$\Delta x \, \Delta p \geq \frac{\hbar}{2}$

where $\hbar = \frac{h}{2\pi}$ is the reduced Planck constant.

This means:

• The more precisely we know a particle's position ($\Delta x$ small), the less precisely we can know its momentum ($\Delta p$ large), and vice versa.
• This is not a limitation of measurement instruments; it is a fundamental property of nature.
• The uncertainty principle has important consequences: electrons cannot orbit the nucleus in classical trajectories (Bohr model is an approximation), and it explains why particles can tunnel through classically forbidden regions.

There is also an energy-time uncertainty relation:

$\Delta E \, \Delta t \geq \frac{\hbar}{2}$

This allows "virtual particles" to briefly exist as long as $\Delta E \, \Delta t \sim \hbar$, and explains the natural width of spectral lines.

1.5 Compton Scattering

When an X-ray photon collides with a free (or loosely bound) electron, the photon is scattered with a longer wavelength. This shift in wavelength is called Compton scattering and provides direct evidence for the particle nature of photons.

The change in wavelength of the scattered photon is:

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

where:

• $\lambda$ is the initial wavelength
• $\lambda'$ is the scattered wavelength
• $\theta$ is the scattering angle of the photon
• $m_e$ is the electron rest mass
• $\frac{h}{m_e c} = 2.43 \times 10^{-12}\,\mathrm{m}$ is the Compton wavelength of the electron

Key observations:

• The wavelength shift depends only on the scattering angle, not the initial wavelength.
• $\Delta\lambda = 0$ when $\theta = 0$ (no scattering) and $\Delta\lambda = \frac{2h}{m_e c}$ when $\theta = 180^\circ$ (back-scattering).
• The scattered photon has less energy than the incident photon; the difference is transferred to the electron as kinetic energy.

Worked Example: Compton Scattering Calculation

An X-ray photon with wavelength $\lambda = 5.00 \times 10^{-12}\,\mathrm{m}$ is scattered at an angle of $\theta = 90^\circ$ by a free electron. Find the wavelength of the scattered photon and the kinetic energy transferred to the electron.

Wavelength shift:

$\Delta\lambda = \frac{h}{m_e c}(1 - \cos 90^\circ) = 2.43 \times 10^{-12}(1 - 0) = 2.43 \times 10^{-12}\,\mathrm{m}$

Scattered wavelength:

$\lambda' = \lambda + \Delta\lambda = 5.00 \times 10^{-12} + 2.43 \times 10^{-12} = 7.43 \times 10^{-12}\,\mathrm{m}$

Energy of incident photon:

$E_i = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{5.00 \times 10^{-12}} = 3.978 \times 10^{-14}\,\mathrm{J}$

Energy of scattered photon:

$E_s = \frac{hc}{\lambda'} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{7.43 \times 10^{-12}} = 2.677 \times 10^{-14}\,\mathrm{J}$

Kinetic energy of electron:

$E_k = E_i - E_s = 3.978 \times 10^{-14} - 2.677 \times 10^{-14} = 1.30 \times 10^{-14}\,\mathrm{J}$

If you get this wrong, revise: Section 1.5.

1.6 Pair Production and Annihilation

Pair production is the process in which a photon is converted into a particle-antiparticle pair, most commonly an electron-positron pair. Energy and momentum must both be conserved.

The minimum photon energy for pair production equals the combined rest energy of the two particles:

$E_{\mathrm{min}} = 2m_e c^2 = 2(9.11 \times 10^{-31})(3.00 \times 10^8)^2 = 1.64 \times 10^{-13}\,\mathrm{J} = 1.02\,\mathrm{MeV}$

In practice, the photon must have more than this energy because momentum conservation requires a nearby nucleus (or another particle) to recoil, absorbing some momentum.

Pair annihilation is the reverse process: when a particle meets its antiparticle, both are converted entirely into photon energy. For example, when an electron and positron annihilate:

$e^- + e^+ \to 2\gamma$

Two photons are produced (not one) to conserve momentum. Each photon has energy $E = m_e c^2 = 0.511\,\mathrm{MeV}$ and the photons travel in opposite directions.

1.7 Quantum Tunnelling (Qualitative)

Quantum tunnelling is the phenomenon where a particle can pass through a potential energy barrier that it classically could not surmount. This is a consequence of the wave nature of particles and the uncertainty principle.

Key points:

• The probability of tunnelling depends on the barrier height and width. Higher and wider barriers have lower transmission probabilities.
• Tunnelling probability increases exponentially as the barrier width decreases.
• The wavefunction does not abruptly go to zero inside the barrier; it decays exponentially.
• There is a non-zero probability of finding the particle on the other side of the barrier.

Applications of quantum tunnelling:

• Alpha decay: An alpha particle inside a nucleus encounters a potential barrier due to the strong nuclear force. Tunnelling through this barrier allows alpha decay.
• Scanning tunnelling microscope (STM): Uses the exponential dependence of tunnelling current on distance to create atomic-resolution images of surfaces.
• Nuclear fusion in stars: Protons tunnel through the Coulomb barrier to undergo fusion at temperatures lower than classical physics would predict.

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2. Nuclear Physics (Topic 7)

2.1 Nuclear Structure

The nucleus of an atom contains protons (positively charged, $+e$) and neutrons (electrically neutral). Together, protons and neutrons are called nucleons.

Key definitions:

• Atomic number (proton number) $Z$: the number of protons in the nucleus.
• Mass number (nucleon number) $A$: the total number of protons and neutrons ($A = Z + N$).
• Neutron number $N = A - Z$: the number of neutrons.
• Nuclide notation: $\prescript{A}{}{Z}\mathrm{X}$, where X is the chemical symbol.
• Isotopes: atoms of the same element with different numbers of neutrons (same $Z$, different $A$).
• Isobars: nuclei with the same mass number but different proton number (same $A$, different $Z$).
• Isotones: nuclei with the same neutron number (same $N$, different $Z$).

The nucleus is extremely small: typical nuclear radius $r \approx 1.2 \times 10^{-15}\,\mathrm{m}$. The empirical formula for nuclear radius is:

$r = r_0 A^{1/3}$

where $r_0 \approx 1.2 \times 10^{-15}\,\mathrm{m}$. The $A^{1/3}$ dependence implies that nuclear density is approximately constant for all nuclei.

Nuclear density:

$\rho = \frac{A \times m_p}{\frac{4}{3}\pi r^3} = \frac{A \times m_p}{\frac{4}{3}\pi (r_0 A^{1/3})^3} = \frac{3m_p}{4\pi r_0^3} \approx 2.3 \times 10^{17}\,\mathrm{kg\,m}^{-3}$

This is enormously greater than typical atomic or everyday densities.

2.2 Radioactive Decay

Unstable nuclei undergo radioactive decay by emitting particles or electromagnetic radiation. There are three main types:

Alpha decay ($\alpha$):

An alpha particle ($\prescript{4}{}{2}\mathrm{He}$, i.e., a helium-4 nucleus) is emitted:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A-4}{}{Z-2}\mathrm{Y} + \prescript{4}{}{2}\mathrm{He}$

Alpha particles are highly ionising but have low penetration (stopped by a few centimetres of air or a sheet of paper).

Beta-minus decay ($\beta^-$):

A neutron converts into a proton, emitting an electron and an electron antineutrino:

$n \to p + e^- + \bar{\nu}_e$

The nuclear equation:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A}{}{Z+1}\mathrm{Y} + e^- + \bar{\nu}_e$

Mass number $A$ is conserved; proton number increases by 1. The antineutrino is required to conserve energy and momentum (the continuous energy spectrum of beta particles led Pauli to propose its existence).

Beta-plus decay ($\beta^+$):

A proton converts into a neutron, emitting a positron and an electron neutrino:

$p \to n + e^+ + \nu_e$

The nuclear equation:

$\prescript{A}{}{Z}\mathrm{X} \to \prescript{A}{}{Z-1}\mathrm{Y} + e^+ + \nu_e$

Beta-plus decay requires the parent nucleus to have sufficient energy to create the positron mass.

Gamma radiation ($\gamma$):

After alpha or beta decay, the daughter nucleus is often left in an excited state. It returns to the ground state by emitting a gamma photon:

$\prescript{A}{}{Z}\mathrm{X}^* \to \prescript{A}{}{Z}\mathrm{X} + \gamma$

Gamma rays are highly penetrating but weakly ionising. They require thick lead or concrete for shielding.

Property	Alpha ($\alpha$)	Beta ($\beta$)	Gamma ($\gamma$)
Particle	$\prescript{4}{}{2}\mathrm{He}$ nucleus	Electron / Positron	Photon
Charge	$+2e$	$\pm e$	0
Ionising power	High	Moderate	Low
Penetrating power	Low	Moderate	High
Stopped by	Paper / few cm air	Aluminium	Lead / concrete

2.3 Half-Life and Decay Constant

Radioactive decay is a random and spontaneous process. The probability that any given nucleus will decay in a given time interval is constant, independent of:

• The number of nuclei present
• External conditions (temperature, pressure, chemical state)
• How long the nucleus has already existed

The decay constant $\lambda$ is the probability per unit time that a nucleus will decay. The number of undecayed nuclei at time $t$ is:

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

where $N_0$ is the initial number of nuclei.

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

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

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

$t_{1/2} = \frac{\ln 2}{\lambda}$

Activity $A$ is the rate of decay (number of decays per unit time):

$A = -\frac{dN}{dt} = \lambda N = A_0 e^{-\lambda t}$

The SI unit of activity is the becquerel (Bq), where $1\,\mathrm{Bq} = 1\,\mathrm{decay/s}$.

Worked Example: Half-Life Calculations

A sample of iodine-131 has a half-life of 8.04 days and an initial activity of $640\,\mathrm{Bq}$.

(a) Find the decay constant.

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.04 \times 24 \times 3600} = \frac{0.693}{694656} = 9.98 \times 10^{-7}\,\mathrm{s}^{-1}$

(b) Find the activity after 2 days.

$A = A_0 e^{-\lambda t} = 640 \times e^{-(9.98 \times 10^{-7})(2 \times 24 \times 3600)}$

$A = 640 \times e^{-0.172} = 640 \times 0.842 = 539\,\mathrm{Bq}$

(c) Find the time for the activity to fall to $40\,\mathrm{Bq}$.

$40 = 640 e^{-\lambda t} \implies e^{-\lambda t} = \frac{40}{640} = 0.0625$

$-\lambda t = \ln 0.0625 = -2.773$

$t = \frac{2.773}{9.98 \times 10^{-7}} = 2.78 \times 10^6\,\mathrm{s} \approx 32.2\,\mathrm{days}$

Check using half-lives: $\frac{640}{40} = 16 = 2^4$, so exactly 4 half-lives = $4 \times 8.04 = 32.2\,\mathrm{days}$.

If you get this wrong, revise: Section 2.3.

2.4 Nuclear Equations and Conservation Laws

All nuclear reactions must satisfy the following conservation laws:

1. Conservation of nucleon number (mass number $A$): the total number of nucleons is the same on both sides.
2. Conservation of proton number (atomic number $Z$): the total charge is the same on both sides.
3. Conservation of energy: total energy (including rest energy) is conserved.
4. Conservation of momentum: total momentum is conserved.

Worked Example: Nuclear Equations

Complete the following nuclear equations:

(a) $\prescript{238}{}{92}\mathrm{U} \to \prescript{234}{}{90}\mathrm{Th} + \prescript{4}{}{2}\mathrm{He}$

This is alpha decay. Check conservation: $A = 238 = 234 + 4$ and $Z = 92 = 90 + 2$.

(b) $\prescript{14}{}{6}\mathrm{C} \to \prescript{14}{}{7}\mathrm{N} + e^- + \bar{\nu}_e$

This is beta-minus decay. Check: $A = 14 = 14 + 0$ and $Z = 6 = 7 + (-1)$.

(c) $\prescript{11}{}{6}\mathrm{C} \to \prescript{11}{}{5}\mathrm{B} + e^+ + \nu_e$

This is beta-plus decay. Check: $A = 11 = 11 + 0$ and $Z = 6 = 5 + 1$.

(d) Identify the unknown particle X in: $\prescript{14}{}{7}\mathrm{N} + \prescript{4}{}{2}\mathrm{He} \to \prescript{17}{}{8}\mathrm{O} + \mathrm{X}$

Conservation of $A$: $14 + 4 = 17 + A_X \implies A_X = 1$

Conservation of $Z$: $7 + 2 = 8 + Z_X \implies Z_X = 1$

Therefore $\mathrm{X} = \prescript{1}{}{1}\mathrm{H}$ (a proton).

If you get this wrong, revise: Section 2.4.

2.5 Binding Energy and Mass Defect

The mass defect $\Delta m$ is the difference between the mass of a nucleus and the total mass of its individual nucleons:

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

This "missing" mass has been converted into the binding energy that holds the nucleus together, given by Einstein's mass-energy equivalence:

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

:::info[Data Booklet Reference] The unified atomic mass unit (u) is defined such that $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$. This conversion factor appears in the IB data booklet.

:::

Binding energy per nucleon is a measure of nuclear stability. Plotting binding energy per nucleon against mass number gives the binding energy curve, which shows:

• Very light nuclei (e.g., $\prescript{2}{}{1}\mathrm{H}$) have low binding energy per nucleon.
• The curve peaks around $A \approx 56$ (iron-56 is the most stable nucleus).
• Heavy nuclei (e.g., uranium) have lower binding energy per nucleon.

The most stable nucleus is $\prescript{62}{}{28}\mathrm{Ni}$ or $\prescript{56}{}{26}\mathrm{Fe}$, with the highest binding energy per nucleon of approximately $8.8\,\mathrm{MeV/nucleon}$.

Worked Example: Binding Energy of Helium-4

Calculate the binding energy per nucleon of $\prescript{4}{}{2}\mathrm{He}$ (alpha particle).

Given masses:

• $m_p = 1.00728\,\mathrm{u}$
• $m_n = 1.00867\,\mathrm{u}$
• $m(\prescript{4}{}{2}\mathrm{He}) = 4.00151\,\mathrm{u}$

Mass defect:

$\Delta m = 2m_p + 2m_n - m_{\mathrm{He}} = 2(1.00728) + 2(1.00867) - 4.00151$

$\Delta m = 2.01456 + 2.01734 - 4.00151 = 0.03039\,\mathrm{u}$

Binding energy:

$E_b = 0.03039 \times 931.5 = 28.3\,\mathrm{MeV}$

Binding energy per nucleon:

$\frac{E_b}{A} = \frac{28.3}{4} = 7.08\,\mathrm{MeV/nucleon}$

If you get this wrong, revise: Section 2.5.

2.6 Nuclear Fission and Fusion

Nuclear fission is the splitting of a heavy nucleus into two (or more) lighter nuclei, accompanied by the release of energy. Fission occurs because the products have a higher binding energy per nucleon than the parent nucleus.

A typical fission reaction:

$\prescript{235}{}{92}\mathrm{U} + \prescript{1}{}{0}\mathrm{n} \to \prescript{236}{}{92}\mathrm{U}^* \to \prescript{141}{}{56}\mathrm{Ba} + \prescript{92}{}{36}\mathrm{Kr} + 3\prescript{1}{}{0}\mathrm{n}$

Key features:

• A chain reaction is possible because each fission event releases neutrons that can trigger further fission events.
• A critical mass is the minimum mass of fissile material needed to sustain a chain reaction.
• Controlled fission occurs in nuclear reactors; uncontrolled fission occurs in nuclear weapons.
• The released energy comes from the mass difference between reactants and products.

Nuclear fusion is the joining of two light nuclei to form a heavier nucleus, releasing energy. Fusion occurs because the product has a higher binding energy per nucleon than the reactants.

Example fusion reactions:

$\prescript{2}{}{1}\mathrm{H} + \prescript{2}{}{1}\mathrm{H} \to \prescript{3}{}{2}\mathrm{He} + \prescript{1}{}{0}\mathrm{n} + 3.27\,\mathrm{MeV}$

$\prescript{2}{}{1}\mathrm{H} + \prescript{3}{}{1}\mathrm{H} \to \prescript{4}{}{2}\mathrm{He} + \prescript{1}{}{0}\mathrm{n} + 17.6\,\mathrm{MeV}$

Fusion requires extremely high temperatures ($\sim 10^7\,\mathrm{K}$) to overcome the electrostatic repulsion between the positively charged nuclei. At these temperatures, matter exists as a plasma.

Conditions for fusion in a reactor:

• High temperature: to give nuclei sufficient kinetic energy to overcome the Coulomb barrier.
• High density: to increase the collision rate between nuclei.
• Confinement time: to keep the plasma contained long enough for sufficient reactions to occur. The Lawson criterion ($n\tau \gt 10^{20}\,\mathrm{s\,m}^{-3}$) gives the minimum product of density and confinement time.

Worked Example: Energy from Fission

Calculate the energy released when $\prescript{235}{}{92}\mathrm{U}$ undergoes fission to produce $\prescript{141}{}{56}\mathrm{Ba}$, $\prescript{92}{}{36}\mathrm{Kr}$, and 3 neutrons.

Given masses:

• $m(\prescript{235}{}{92}\mathrm{U}) = 235.044\,\mathrm{u}$
• $m(\prescript{141}{}{56}\mathrm{Ba}) = 140.914\,\mathrm{u}$
• $m(\prescript{92}{}{36}\mathrm{Kr}) = 91.926\,\mathrm{u}$
• $m_n = 1.009\,\mathrm{u}$

Mass of reactants: $235.044 + 1.009 = 236.053\,\mathrm{u}$

Mass of products: $140.914 + 91.926 + 3(1.009) = 235.867\,\mathrm{u}$

Mass defect:

$\Delta m = 236.053 - 235.867 = 0.186\,\mathrm{u}$

Energy released:

$E = 0.186 \times 931.5 = 173\,\mathrm{MeV}$

This energy appears as kinetic energy of the fission fragments, neutrons, and gamma radiation.

If you get this wrong, revise: Section 2.6.

2.7 Background Radiation and Detection

Background radiation is the ionising radiation that is always present in the environment. Sources include:

• Cosmic rays: high-energy particles from space (primarily protons and alpha particles).
• Terrestrial radiation: radioactive isotopes in rocks and soil (e.g., uranium, thorium, radon-222 gas).
• Medical sources: X-rays, radioactive tracers used in diagnosis and treatment.
• Nuclear power and weapons testing: fallout and waste from nuclear facilities.

Typical background radiation dose is approximately $2.4\,\mathrm{mSv}$ per year, varying by location.

Radiation detection methods:

• Geiger-Muller (GM) tube: detects ionising radiation by the ionisation of gas inside the tube. Each particle produces a current pulse, counted electronically. It does not distinguish between types of radiation.
• Photographic film: darkens when exposed to radiation. Used in radiation badges worn by workers.
• Scintillation detector: a material (e.g., sodium iodide) that emits light when radiation passes through it. The light is converted to an electrical signal by a photomultiplier tube.
• Semiconductor detector: radiation creates electron-hole pairs in a semiconductor material, producing a current proportional to the energy deposited.

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3. Worked Examples

Worked Example 1: Photoelectric Effect

Light of wavelength $400\,\mathrm{nm}$ is incident on a sodium surface with work function $\phi = 2.28\,\mathrm{eV}$.

(a) Calculate the maximum kinetic energy of emitted photoelectrons.

Photon energy:

$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19}\,\mathrm{J}$

Convert to eV: $E = \frac{4.97 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.11\,\mathrm{eV}$

Maximum kinetic energy:

$E_{k,\mathrm{max}} = E - \phi = 3.11 - 2.28 = 0.83\,\mathrm{eV}$

(b) Calculate the stopping potential.

$V_s = \frac{E_{k,\mathrm{max}}}{e} = \frac{0.83\,\mathrm{eV}}{e} = 0.83\,\mathrm{V}$

(c) Calculate the threshold frequency.

$f_0 = \frac{\phi}{h} = \frac{2.28 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.50 \times 10^{14}\,\mathrm{Hz}$

Worked Example 2: De Broglie Wavelength

An electron is accelerated through a potential difference of $200\,\mathrm{V}$. Calculate its de Broglie wavelength.

The kinetic energy gained by the electron:

$E_k = eV = (1.60 \times 10^{-19})(200) = 3.20 \times 10^{-17}\,\mathrm{J}$

Since $E_k = \frac{p^2}{2m_e}$, the momentum is:

$p = \sqrt{2m_e E_k} = \sqrt{2(9.11 \times 10^{-31})(3.20 \times 10^{-17})} = \sqrt{5.83 \times 10^{-47}} = 7.64 \times 10^{-24}\,\mathrm{kg\,m\,s}^{-1}$

De Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{7.64 \times 10^{-24}} = 8.68 \times 10^{-11}\,\mathrm{m}$

This is comparable to atomic spacings, explaining why electron diffraction can probe crystal structures.

Worked Example 3: Pair Production

A photon with energy $3.00\,\mathrm{MeV}$ undergoes pair production, creating an electron-positron pair. Calculate the total kinetic energy of the pair.

The rest energy of an electron (or positron) is:

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

Total rest energy of the pair: $2 \times 0.511 = 1.022\,\mathrm{MeV}$

Total kinetic energy:

$E_k = E_{\mathrm{photon}} - 2m_e c^2 = 3.00 - 1.022 = 1.978\,\mathrm{MeV}$

This kinetic energy is shared between the electron and positron (not necessarily equally, depending on how momentum is shared with any recoiling nucleus).

Worked Example 4: Nuclear Binding Energy

Calculate the binding energy per nucleon of $\prescript{56}{}{26}\mathrm{Fe}$.

Given masses:

• $m_p = 1.00728\,\mathrm{u}$
• $m_n = 1.00867\,\mathrm{u}$
• $m(\prescript{56}{}{26}\mathrm{Fe}) = 55.93493\,\mathrm{u}$

Number of protons: $Z = 26$, number of neutrons: $N = 56 - 26 = 30$.

Mass defect:

$\Delta m = 26(1.00728) + 30(1.00867) - 55.93493$

$\Delta m = 26.18928 + 30.26010 - 55.93493 = 0.51445\,\mathrm{u}$

Binding energy:

$E_b = 0.51445 \times 931.5 = 479.2\,\mathrm{MeV}$

Binding energy per nucleon:

$\frac{E_b}{A} = \frac{479.2}{56} = 8.56\,\mathrm{MeV/nucleon}$

This is close to the maximum on the binding energy curve, confirming that iron-56 is among the most stable nuclei.

Worked Example 5: Radioactive Dating

A sample of ancient wood contains $25\%$ of the carbon-14 found in a living tree. The half-life of carbon-14 is 5730 years. Estimate the age of the sample.

The fraction remaining is $\frac{N}{N_0} = 0.25 = \frac{1}{4} = 2^{-2}$.

Since the fraction is $\left(\frac{1}{2}\right)^n$ where $n$ is the number of half-lives, we have $n = 2$ half-lives.

Age $= 2 \times 5730 = 11460\,\mathrm{years}$.

Using the exponential formula:

$N = N_0 e^{-\lambda t} \implies 0.25 = e^{-\lambda t}$

$\lambda t = \ln 4 = 1.386$

$t = \frac{1.386}{\lambda} = \frac{1.386 \times 5730}{\ln 2} = \frac{1.386 \times 5730}{0.693} = 11460\,\mathrm{years}$

Worked Example 6: Heisenberg Uncertainty Principle

An electron is confined to a region of width $\Delta x = 1.0 \times 10^{-10}\,\mathrm{m}$ (roughly the size of an atom). Estimate the minimum uncertainty in its momentum.

Using the Heisenberg uncertainty principle:

$\Delta x \, \Delta p \geq \frac{\hbar}{2} = \frac{h}{4\pi}$

$\Delta p \geq \frac{h}{4\pi \, \Delta x} = \frac{6.63 \times 10^{-34}}{4\pi(1.0 \times 10^{-10})} = 5.27 \times 10^{-25}\,\mathrm{kg\,m\,s}^{-1}$

The corresponding minimum uncertainty in velocity:

$\Delta v = \frac{\Delta p}{m_e} = \frac{5.27 \times 10^{-25}}{9.11 \times 10^{-31}} = 5.79 \times 10^5\,\mathrm{m/s}$

This is a significant fraction of the speed of light, illustrating that confinement to atomic dimensions leads to large momentum uncertainties, consistent with the quantum behaviour of electrons in atoms.

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Summary of Key Results

Concept	Key Formula
Photon energy	$E = hf$
Photoelectric equation	$E_{k,\mathrm{max}} = hf - \phi$
Work function	$\phi = hf_0$
De Broglie wavelength	$\lambda = \frac{h}{p}$
Heisenberg uncertainty	$\Delta x \, \Delta p \geq \frac{\hbar}{2}$
Compton scattering	$\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)$
Energy levels (hydrogen)	$E_n = \frac{-13.6}{n^2}\,\mathrm{eV}$
Pair production minimum	$E_{\mathrm{min}} = 2m_e c^2$
Radioactive decay	$N = N_0 e^{-\lambda t}$
Half-life	$t_{1/2} = \frac{\ln 2}{\lambda}$
Activity	$A = \lambda N$
Binding energy	$E_b = \Delta m \, c^2$
Nuclear radius	$r = r_0 A^{1/3}$
Mass-energy conversion	$1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$

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Problem Set

Problem 1

Light of wavelength $550\,\mathrm{nm}$ is incident on a metal surface. The stopping potential is found to be $0.45\,\mathrm{V}$. Calculate the work function of the metal in eV.

Solution

Photon energy:

$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{550 \times 10^{-9}} = 3.62 \times 10^{-19}\,\mathrm{J} = 2.26\,\mathrm{eV}$

Using $E_{k,\mathrm{max}} = eV_s = 0.45\,\mathrm{eV}$:

$\phi = E - E_{k,\mathrm{max}} = 2.26 - 0.45 = 1.81\,\mathrm{eV}$

If you get this wrong, revise: Section 1.1.

Problem 2

The threshold wavelength for a certain metal is $300\,\mathrm{nm}$. Calculate the maximum kinetic energy of photoelectrons when light of wavelength $200\,\mathrm{nm}$ is incident on the surface.

Solution

Work function:

$\phi = \frac{hc}{\lambda_0} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{300 \times 10^{-9}} = 6.63 \times 10^{-19}\,\mathrm{J} = 4.14\,\mathrm{eV}$

Photon energy at $200\,\mathrm{nm}$:

$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{200 \times 10^{-9}} = 9.945 \times 10^{-19}\,\mathrm{J} = 6.22\,\mathrm{eV}$

Maximum kinetic energy:

$E_{k,\mathrm{max}} = 6.22 - 4.14 = 2.08\,\mathrm{eV}$

If you get this wrong, revise: Section 1.1.

Problem 3

A neutron moving with velocity $2.0 \times 10^4\,\mathrm{m/s}$ has a de Broglie wavelength of $1.98 \times 10^{-11}\,\mathrm{m}$. Calculate the mass of the neutron from this data and verify it agrees with the accepted value of $1.675 \times 10^{-27}\,\mathrm{kg}$.

Solution

Using $\lambda = \frac{h}{mv}$:

$m = \frac{h}{\lambda v} = \frac{6.63 \times 10^{-34}}{(1.98 \times 10^{-11})(2.0 \times 10^4)} = \frac{6.63 \times 10^{-34}}{3.96 \times 10^{-7}} = 1.67 \times 10^{-27}\,\mathrm{kg}$

This agrees with the accepted value of $1.675 \times 10^{-27}\,\mathrm{kg}$ to three significant figures.

If you get this wrong, revise: Section 1.2.

Problem 4

An electron in a hydrogen atom is in the $n = 3$ energy level. Calculate the wavelengths of all possible photons that could be emitted as the electron returns to the ground state.

Solution

Possible transitions: $3 \to 2$, $3 \to 1$, and $2 \to 1$ (after first transitioning to $n = 2$).

For $3 \to 2$ (Balmer series):

$\Delta E = \frac{-13.6}{9} - \frac{-13.6}{4} = -1.51 - (-3.40) = 1.89\,\mathrm{eV}$

$\lambda = \frac{hc}{\Delta E} = \frac{1240\,\mathrm{eV\,nm}}{1.89\,\mathrm{eV}} = 656\,\mathrm{nm}$

For $3 \to 1$ (Lyman series):

$\Delta E = \frac{-13.6}{9} - \frac{-13.6}{1} = -1.51 - (-13.6) = 12.09\,\mathrm{eV}$

$\lambda = \frac{1240}{12.09} = 102.6\,\mathrm{nm}$

For $2 \to 1$ (Lyman series):

$\Delta E = \frac{-13.6}{4} - \frac{-13.6}{1} = -3.40 - (-13.6) = 10.2\,\mathrm{eV}$

$\lambda = \frac{1240}{10.2} = 121.6\,\mathrm{nm}$

The three possible wavelengths are $656\,\mathrm{nm}$, $121.6\,\mathrm{nm}$, and $102.6\,\mathrm{nm}$.

If you get this wrong, revise: Section 1.3.

Problem 5

A proton is confined within a nucleus of radius $5.0 \times 10^{-15}\,\mathrm{m}$. Estimate the minimum uncertainty in its kinetic energy.

Solution

Using the uncertainty principle with $\Delta x \approx 2r = 1.0 \times 10^{-14}\,\mathrm{m}$:

$\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2(1.0 \times 10^{-14})} = 5.28 \times 10^{-21}\,\mathrm{kg\,m\,s}^{-1}$

For a non-relativistic proton ($m_p = 1.67 \times 10^{-27}\,\mathrm{kg}$):

$E_k \geq \frac{(\Delta p)^2}{2m_p} = \frac{(5.28 \times 10^{-21})^2}{2(1.67 \times 10^{-27})} = \frac{2.79 \times 10^{-41}}{3.34 \times 10^{-27}} = 8.35 \times 10^{-15}\,\mathrm{J}$

$E_k \geq \frac{8.35 \times 10^{-15}}{1.60 \times 10^{-19}} = 5.22 \times 10^4\,\mathrm{eV} = 52.2\,\mathrm{keV}$

If you get this wrong, revise: Section 1.4.

Problem 6

X-rays of wavelength $1.00 \times 10^{-11}\,\mathrm{m}$ are scattered at an angle of $60^\circ$ by free electrons. Calculate the wavelength of the scattered X-rays and the kinetic energy of the recoil electrons.

Solution

Wavelength shift:

$\Delta\lambda = \frac{h}{m_e c}(1 - \cos 60^\circ) = 2.43 \times 10^{-12}(1 - 0.5) = 1.215 \times 10^{-12}\,\mathrm{m}$

Scattered wavelength:

$\lambda' = 1.00 \times 10^{-11} + 1.215 \times 10^{-12} = 1.12 \times 10^{-11}\,\mathrm{m}$

Energy of incident photon:

$E_i = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{0.0100\,\mathrm{nm}} = 1.24 \times 10^5\,\mathrm{eV} = 124\,\mathrm{keV}$

Energy of scattered photon:

$E_s = \frac{hc}{\lambda'} = \frac{1240}{0.0112} = 1.107 \times 10^5\,\mathrm{eV} = 110.7\,\mathrm{keV}$

Kinetic energy of recoil electron:

$E_k = E_i - E_s = 124 - 110.7 = 13.3\,\mathrm{keV}$

If you get this wrong, revise: Section 1.5.

Problem 7

A radioactive isotope has a half-life of 5.27 years and an initial activity of $800\,\mathrm{Bq}$. Calculate the time taken for the activity to decrease to $50\,\mathrm{Bq}$.

Solution

Number of half-lives: $\frac{800}{50} = 16 = 2^4$, so $n = 4$ half-lives.

Time $= 4 \times 5.27 = 21.1\,\mathrm{years}$.

Using the exponential formula:

$\lambda = \frac{\ln 2}{5.27 \times 365.25 \times 24 \times 3600} = 4.17 \times 10^{-9}\,\mathrm{s}^{-1}$

$50 = 800 e^{-\lambda t} \implies t = \frac{\ln 16}{\lambda} = \frac{2.773}{4.17 \times 10^{-9}} = 6.65 \times 10^8\,\mathrm{s} \approx 21.1\,\mathrm{years}$

If you get this wrong, revise: Section 2.3.

Problem 8

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

$\prescript{226}{}{88}\mathrm{Ra} \to \prescript{222}{}{86}\mathrm{Rn} + \prescript{A}{}{Z}\mathrm{X}$

Also calculate the energy released given the following masses: $m(\prescript{226}{}{88}\mathrm{Ra}) = 226.02540\,\mathrm{u}$, $m(\prescript{222}{}{86}\mathrm{Rn}) = 222.01757\,\mathrm{u}$, $m(\prescript{4}{}{2}\mathrm{He}) = 4.00260\,\mathrm{u}$.

Solution

Conservation of $A$: $226 = 222 + A \implies A = 4$

Conservation of $Z$: $88 = 86 + Z \implies Z = 2$

So $\mathrm{X} = \prescript{4}{}{2}\mathrm{He}$ (alpha particle). This is alpha decay.

Mass defect:

$\Delta m = 226.02540 - (222.01757 + 4.00260) = 226.02540 - 226.02017 = 0.00523\,\mathrm{u}$

Energy released:

$E = 0.00523 \times 931.5 = 4.87\,\mathrm{MeV}$

If you get this wrong, revise: Sections 2.2 and 2.4.

Problem 9

Calculate the binding energy per nucleon of lithium-7 ($\prescript{7}{}{3}\mathrm{Li}$).

Given: $m(\prescript{7}{}{3}\mathrm{Li}) = 7.01600\,\mathrm{u}$, $m_p = 1.00728\,\mathrm{u}$, $m_n = 1.00867\,\mathrm{u}$.

Solution

$Z = 3$ protons, $N = 7 - 3 = 4$ neutrons.

Mass defect:

$\Delta m = 3(1.00728) + 4(1.00867) - 7.01600$

$\Delta m = 3.02184 + 4.03468 - 7.01600 = 0.04052\,\mathrm{u}$

Binding energy:

$E_b = 0.04052 \times 931.5 = 37.7\,\mathrm{MeV}$

Binding energy per nucleon:

$\frac{E_b}{A} = \frac{37.7}{7} = 5.39\,\mathrm{MeV/nucleon}$

If you get this wrong, revise: Section 2.5.

Problem 10

In a nuclear fusion reaction, two deuterium nuclei ($\prescript{2}{}{1}\mathrm{H}$) fuse to form helium-3 ($\prescript{3}{}{2}\mathrm{He}$) and a neutron. Calculate the energy released.

Given: $m(\prescript{2}{}{1}\mathrm{H}) = 2.01410\,\mathrm{u}$, $m(\prescript{3}{}{2}\mathrm{He}) = 3.01603\,\mathrm{u}$, $m_n = 1.00867\,\mathrm{u}$.

Solution

Reaction: $\prescript{2}{}{1}\mathrm{H} + \prescript{2}{}{1}\mathrm{H} \to \prescript{3}{}{2}\mathrm{He} + \prescript{1}{}{0}\mathrm{n}$

Mass of reactants: $2(2.01410) = 4.02820\,\mathrm{u}$

Mass of products: $3.01603 + 1.00867 = 4.02470\,\mathrm{u}$

Mass defect:

$\Delta m = 4.02820 - 4.02470 = 0.00350\,\mathrm{u}$

Energy released:

$E = 0.00350 \times 931.5 = 3.26\,\mathrm{MeV}$

If you get this wrong, revise: Section 2.6.

Problem 11

A GM tube records a count rate of $120\,\mathrm{counts/min}$ from a radioactive source. When a $2.0\,\mathrm{mm}$ thick aluminium sheet is placed in front of the source, the count rate drops to $15\,\mathrm{counts/min}$. When the sheet is replaced by a $5.0\,\mathrm{mm}$ thick lead sheet, the count rate drops to $10\,\mathrm{counts/min}$. The background count rate is $10\,\mathrm{counts/min}$. Identify the types of radiation emitted by the source.

Solution

Corrected count rate without absorber: $120 - 10 = 110\,\mathrm{counts/min}$

With aluminium: $15 - 10 = 5\,\mathrm{counts/min}$ (almost all radiation stopped)

With lead: $10 - 10 = 0\,\mathrm{counts/min}$ (all radiation stopped)

The aluminium stops most of the radiation, indicating the presence of alpha radiation (which is stopped by thin aluminium or even paper). The small residual count rate with aluminium that is fully eliminated by lead indicates a small component of beta radiation.

No significant gamma component is detected since the lead stops everything.

The source emits alpha and beta radiation.

If you get this wrong, revise: Sections 2.2 and 2.7.

Problem 12

A positron and an electron, each with negligible kinetic energy, annihilate to produce two gamma-ray photons. Calculate the wavelength of each photon.

Solution

The total energy of the electron-positron pair is their combined rest energy:

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

This energy is shared equally between the two photons (for zero initial kinetic energy):

$E_\gamma = 0.511\,\mathrm{MeV} = 8.18 \times 10^{-14}\,\mathrm{J}$

Wavelength of each photon:

$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{8.18 \times 10^{-14}} = 2.43 \times 10^{-12}\,\mathrm{m}$

Note that this equals the Compton wavelength of the electron, $\frac{h}{m_e c}$.

If you get this wrong, revise: Section 1.6.

For the A-Level treatment of this topic, see Radioactivity.