Quantum and Nuclear Physics | IB

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.

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.

At the threshold, $E_k = 0$:

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

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

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

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.

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

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

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 $500\,\mathrm{nm}$:

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

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.

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.

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 $\lambda = h/(mv)$.

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.

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:

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

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

• $n = 1$: 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 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, $n_i \gt n_f$.

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 ($\mathrm{H}\alpha$).

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.

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.

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: $n = 15.81 / 5.27 = 3$.

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

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.

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.

$E = mc^2$

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.

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.

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.

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} $1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$.

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

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

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

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

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

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

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).

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

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

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.

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.

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.

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

• 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.

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.

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):

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

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.

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.

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.

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

• 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.

$e^- + e^+ \to \gamma + \gamma$: 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.

From Newton’s apple to quantum particles, physics explains how the world works through measurable quantities and testable laws. Forces cause acceleration, energy transforms between forms but is never lost, and waves carry information across vast distances. These concepts form the foundation for engineering, astronomy, and our understanding of reality itself.

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