Level 5 — MasteryModern Physics

Modern Physics

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Time limit: 90 minutes Total marks: 60 Instructions: Answer all THREE questions. Full derivations required. Constants: $h=6.626\times10^{-34}\,\text{J·s}$ , $\hbar=1.055\times10^{-34}\,\text{J·s}$ , $c=3.00\times10^8\,\text{m/s}$ , $m_e=9.11\times10^{-31}\,\text{kg}$ , $1\,\text{eV}=1.602\times10^{-19}\,\text{J}$ , electron rest energy $=0.511\,\text{MeV}$ .

Question 1 — Compton scattering meets relativistic kinematics (20 marks)

An X-ray photon of wavelength $\lambda_0 = 2.00\times10^{-11}\,\text{m}$ scatters off a free electron initially at rest.

(a) Starting from conservation of relativistic energy and momentum, derive the Compton wavelength-shift formula $\Delta\lambda = \frac{h}{m_e c}(1-\cos\theta).$ Clearly state where you use $E^2=(pc)^2+(mc^2)^2$ and $E_\gamma = pc$ . (8)

(b) For scattering angle $\theta = 90°$ , compute the scattered wavelength $\lambda'$ , and the kinetic energy (in keV) handed to the recoiling electron. (6)

(c) Show that a free electron cannot fully absorb the photon (i.e. Compton scattering, not absorption, must occur). Argue from simultaneous conservation of energy and momentum. (4)

(d) Explain in one or two sentences why visible light shows a negligible Compton shift while X-rays show a measurable one, referencing the relevant length scale. (2)

Question 2 — Particle in a box: build, prove, and code (22 marks)

Consider a particle of mass $m$ confined to a 1-D infinite square well of width $L$ , $V=0$ for $0<x<L$ and $V=\infty$ otherwise.

(a) Solve the time-independent Schrödinger equation to derive the normalized eigenfunctions $\psi_n(x)$ and energies $E_n$ . Justify the boundary conditions from continuity of $\psi$ and the normalization of $|\psi|^2$ . (8)

(b) For an electron with $L = 0.50\,\text{nm}$ , compute $E_1$ and $E_2$ in eV, and the wavelength of the photon emitted in the $2\to1$ transition. (5)

(c) Prove the orthogonality relation $\int_0^L \psi_m^*\psi_n\,dx = \delta_{mn}$ for this well. (4)

(d) Write a short (pseudo)code / algorithm using numpy that verifies numerically (i) normalization of $\psi_1$ and (ii) that $E_2/E_1 = 4$ . State what numerical quantities you would print and their expected values. (3)

(e) Using the uncertainty principle $\Delta x\,\Delta p \gtrsim \hbar/2$ , estimate a lower-bound ground-state energy and compare its order of magnitude to the exact $E_1$ . (2)

Question 3 — Nuclear energetics and relativity crossover (18 marks)

(a) For the deuterium–tritium fusion reaction $^2_1\text{H} + \,^3_1\text{H} \rightarrow \,^4_2\text{He} + \,^1_0\text{n},$ given atomic masses (u): $^2\text{H}=2.014102$ , $^3\text{H}=3.016049$ , $^4\text{He}=4.002602$ , neutron $=1.008665$ , and $1\,\text{u}=931.5\,\text{MeV}/c^2$ , compute the Q-value in MeV. (5)

(b) The released energy appears mostly as kinetic energy of the neutron and the $\alpha$ -particle. Using non-relativistic momentum conservation (products from rest), find the fraction of $Q$ carried by the neutron. Comment on whether a non-relativistic treatment is justified. (6)

(c) A $^{4}\text{He}$ nucleus produced above is accelerated to $v=0.60c$ . Using $E^2=(pc)^2+(mc^2)^2$ with rest energy $3727\,\text{MeV}$ , compute its relativistic momentum $p$ (in MeV/ $c$ ) and total energy. (5)

(d) Briefly explain, via the binding-energy-per-nucleon curve, why fusion of light nuclei and fission of heavy nuclei both release energy. (2)

End of paper

Answer keyMark scheme & solutions

Question 1

(a) Derivation (8)

Conservation of energy: $E_0 + m_ec^2 = E' + E_e$ where $E_0=hc/\lambda_0$ , $E'=hc/\lambda'$ (photon energies), $E_e$ total electron energy. (1)
Conservation of momentum (vector): $\vec p_0 = \vec p' + \vec p_e$ , using $p_\gamma=E_\gamma/c=h/\lambda$ . (1)
Square momentum: $p_e^2 = p_0^2 + p'^2 - 2p_0p'\cos\theta$ . (1)
Electron relation $E_e^2 = (p_e c)^2 + (m_ec^2)^2$ ; and $E_e = E_0-E'+m_ec^2$ . (1)
Expand: $(E_0-E'+m_ec^2)^2 = (p_ec)^2 + (m_ec^2)^2$ . Using $E_\gamma=p_\gamma c$ : $E_0=p_0c$ , $E'=p'c$ . (1)
Left side expands to $E_0^2 + E'^2 + m^2c^4 -2E_0E' +2m_ec^2(E_0-E')$ . Right side $= (p_0^2+p'^2-2p_0p'\cos\theta)c^2 + m^2c^4$ . Substituting $p_ic=E_i$ cancels $E_0^2, E'^2$ terms. (1)
Leaves $-2E_0E' + 2m_ec^2(E_0-E') = -2E_0E'\cos\theta$ , i.e. $m_ec^2(E_0-E')=E_0E'(1-\cos\theta)$ . (1)
Divide by $E_0E'/hc$ ... using $E=hc/\lambda$ : $m_ec^2\,hc(1/\lambda'-1/\lambda_0)/...$ → $\lambda'-\lambda_0 = \frac{h}{m_ec}(1-\cos\theta)$ . (1)

(b) Numeric (6)

Compton wavelength $\lambda_C = h/(m_ec) = 6.626\times10^{-34}/(9.11\times10^{-31}\times3\times10^8) = 2.426\times10^{-12}\,\text{m}$ . (1)
$\theta=90°$ : $\Delta\lambda=\lambda_C(1-0)=2.426\times10^{-12}\,\text{m}$ . (1)
$\lambda' = 2.00\times10^{-11}+0.2426\times10^{-11} = 2.243\times10^{-11}\,\text{m}$ . (1)
$E_0 = hc/\lambda_0 = (1240\,\text{eV·nm})/0.0200\,\text{nm} = 62.0\,\text{keV}$ . (1)
$E' = 1240/0.02243 = 55.3\,\text{keV}$ . (1)
$K_e = E_0-E' = 62.0-55.3 = 6.7\,\text{keV}$ . (1)

(c) Free electron cannot absorb (4)

Assume absorption: $hf + m_ec^2 = E_e$ and $h/\lambda = p_e$ . (1)
Then $E_e = hf+m_ec^2$ and $p_ec = hf$ → $E_e - p_ec = m_ec^2$ . (1)
But $E_e^2-(p_ec)^2 = (m_ec^2)^2$ ⟹ $(E_e-p_ec)(E_e+p_ec)=(m_ec^2)^2$ . Substituting $E_e-p_ec=m_ec^2$ gives $E_e+p_ec=m_ec^2$ , so $p_ec=0$ ⟹ $hf=0$ . (1)
Contradiction (no photon). Hence a free electron cannot absorb a photon; a third body (scattered photon) is required — Compton scattering. (1)

(d) (2) The shift $\Delta\lambda\sim\lambda_C=2.4\times10^{-12}$ m is fixed; for visible light ( $\lambda\sim500$ nm) the fractional shift $\Delta\lambda/\lambda\sim10^{-5}$ is negligible, whereas for X-rays ( $\lambda\sim0.01$ nm) it is comparable to $\lambda$ and measurable. (2)

Question 2

(a) Solve TISE (8)

Inside: $-\frac{\hbar^2}{2m}\psi''=E\psi$ ⟹ $\psi''=-k^2\psi$ , $k=\sqrt{2mE}/\hbar$ . (2)
General $\psi=A\sin kx+B\cos kx$ . BC $\psi(0)=0$ (continuity, $\psi=0$ outside) ⟹ $B=0$ . (1)
$\psi(L)=0$ ⟹ $\sin kL=0$ ⟹ $kL=n\pi$ , $n=1,2,\dots$ . (1)
$E_n=\frac{\hbar^2k^2}{2m}=\frac{n^2\pi^2\hbar^2}{2mL^2}=\frac{n^2h^2}{8mL^2}$ . (2)
Normalize: $\int_0^L A^2\sin^2(n\pi x/L)dx=A^2L/2=1$ ⟹ $A=\sqrt{2/L}$ . So $\psi_n=\sqrt{2/L}\sin(n\pi x/L)$ . (2)

(b) Numeric (5)

$E_1 = \frac{h^2}{8mL^2}$ . With $L=0.5\times10^{-9}$ : $E_1=\frac{(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(0.5\times10^{-9})^2}$ . (1)
Numerator $=4.39\times10^{-67}$ ; denominator $=8\times9.11\times10^{-31}\times2.5\times10^{-19}=1.822\times10^{-48}$ . $E_1=2.41\times10^{-19}\,\text{J}=1.505\,\text{eV}$ . (2)
$E_2=4E_1=6.02\,\text{eV}$ . (1)
Photon $\Delta E = E_2-E_1=3E_1=4.51\,\text{eV}$ ; $\lambda=1240/4.51=275\,\text{nm}$ . (1)

(c) Orthogonality (4)

$\int_0^L \frac{2}{L}\sin\frac{m\pi x}{L}\sin\frac{n\pi x}{L}dx$ . (1)
Product-to-sum: $=\frac{1}{L}\int_0^L[\cos\frac{(m-n)\pi x}{L}-\cos\frac{(m+n)\pi x}{L}]dx$ . (1)
For $m\neq n$ both integrals of full cosines over integer periods vanish ⟹ 0. (1)
For $m=n$ : first term $\cos0=1$ integrates to $L$ , second vanishes ⟹ value $=1$ . Hence $\delta_{mn}$ . (1)

(d) Code (3)

import numpy as np
L=1.0; N=100000
x=np.linspace(0,L,N)
def psi(n): return np.sqrt(2/L)*np.sin(n*np.pi*x/L)
norm=np.trapz(psi(1)**2,x)          # expect ~1.0
E=lambda n: n**2                    # E_n ∝ n^2 (units of h^2/8mL^2)
print(norm, E(2)/E(1))              # prints ~1.0 and 4.0

Print norm≈1.000 and E2/E1=4.0. (3)

(e) Uncertainty estimate (2)

$\Delta x\sim L$ , $\Delta p\gtrsim\hbar/(2L)$ , $E\sim\frac{(\Delta p)^2}{2m}=\frac{\hbar^2}{8mL^2}$ . Exact $E_1=\frac{\pi^2\hbar^2}{2mL^2}$ ; ratio $\sim\pi^2/4\approx2.5$ — same order of magnitude. (2)

Question 3

(a) Q-value (5)

$\Delta m = (2.014102+3.016049)-(4.002602+1.008665)$ . (2)
Reactants $=5.030151$ ; products $=5.011267$ . $\Delta m=0.018884\,\text{u}$ . (2)
$Q=0.018884\times931.5=17.59\,\text{MeV}$ . (1)

(b) Neutron energy fraction (6)

Momentum conservation (rest): $p_n=p_\alpha=p$ . (1)
$K=p^2/2m$ ; $K_n/K_\alpha = m_\alpha/m_n$ . (1)
$K_n = Q\cdot\frac{m_\alpha}{m_\alpha+m_n} = Q\cdot\frac{4.0026}{5.0113}=0.799Q$ . (2)
Fraction $\approx0.80$ ⟹ $K_n\approx14.1\,\text{MeV}$ . (1)
Justification: $K_n=14.1$ MeV vs neutron rest energy 939.6 MeV; $K/mc^2\approx0.015\ll1$ , so non-relativistic treatment is acceptable. (1)

(c) Relativistic α (5)

$\gamma=1/\sqrt{1-0.36}=1/\sqrt{0.64}=1.25$ . (1)
$E=\gamma mc^2=1.25\times3727=4658.75\,\text{MeV}$ . (1)
$p c=\sqrt{E^2-(mc^2)^2}=\sqrt{4658.75^2-3727^2}$ . (1)
$=\sqrt{21703...-13890...}\times... = \sqrt{7813... }\;\text{MeV} \approx 2796\,\text{MeV}$ ; so $p\approx2796\,\text{MeV}/c$ . (1)
Check $p=\gamma m v/c\cdot c=\gamma(mc^2)(v/c)=1.25\times3727\times0.6=2795\,\text{MeV}/c$ ✓. (1)

(d) (2) BE-per-nucleon peaks near $^{56}$ Fe. Fusing light nuclei moves toward the peak (higher BE/nucleon ⟹ energy released); fissioning heavy nuclei also moves toward the peak from the other side. Both increase total binding energy, releasing the difference. (2)

[
 {"claim":"Compton shift at 90deg equals Compton wavelength ~2.426e-12 m","code":"h=6.626e-34; me=9.11e-31; c=3.0e8; dl=h/(me*c); result = abs(dl-2.426e-12) < 5e-15"},
 {"claim":"Ground state energy of electron in 0.5nm box ~1.505 eV","code":"h=6.626e-34; me=9.11e-31; L=0.5e-9; E1=h**2/(8*me*L**2)/1.602e-19; result = abs(E1-1.505) < 0.05"},
 {"claim":"E2/E1 = 4 for particle in box","code":"result = (2**2)/(1**2) == 4"},
 {"claim":"DT fusion Q-value ~17.59 MeV","code":"dm=(2.014102+3.016049)-(4.002602+1.008665); Q=dm*931.5; result = abs(Q-17.59) < 0.1"},
 {"claim":"Neutron carries ~0.799 of Q","code":"ma=4.0026; mn=1.008665; frac=ma/(ma+mn); result = abs(frac-0.799) < 0.005"},
 {"claim":"Relativistic alpha momentum at 0.6c ~2795 MeV/c","code":"import sympy as sp; g=1/sp.sqrt(1-0.36); E=g*3727; p=sp.sqrt(E**2-3727**2); result = abs(float(p)-2795) < 5"}
]