Modern Physics

Q1 — Compton scattering (12)

(a) Conservation (3):

• Energy: $\dfrac{hc}{\lambda} + m_ec^2 = \dfrac{hc}{\lambda'} + E_e$ where $E_e=\sqrt{(p_ec)^2+(m_ec^2)^2}$. (1)
• Momentum x: $\dfrac{h}{\lambda} = \dfrac{h}{\lambda'}\cos\theta + p_e\cos\phi$. (1)
• Momentum y: $0 = \dfrac{h}{\lambda'}\sin\theta - p_e\sin\phi$. (1)

(b) Derivation (6): Eliminate $\phi$: from momentum eqs, $p_e^2 c^2 = \left(\frac{hc}{\lambda}\right)^2 + \left(\frac{hc}{\lambda'}\right)^2 - 2\frac{h^2c^2}{\lambda\lambda'}\cos\theta.$ (2) From energy: $E_e = hc/\lambda - hc/\lambda' + m_ec^2$; square and use $E_e^2 = p_e^2c^2 + m_e^2c^4$: $\left(\frac{hc}{\lambda}-\frac{hc}{\lambda'}\right)^2 + 2m_ec^2\left(\frac{hc}{\lambda}-\frac{hc}{\lambda'}\right)+m_e^2c^4 = p_e^2c^2 + m_e^2c^4.$ (2) Substitute $p_e^2c^2$, cancel, expand. The $(hc/\lambda-hc/\lambda')^2$ terms cancel against the momentum expansion leaving: $2m_ec^2\,hc\left(\frac1\lambda-\frac1{\lambda'}\right) = 2\frac{h^2c^2}{\lambda\lambda'}(1-\cos\theta).$ Divide by $2m_ec^2 hc/(\lambda\lambda')$: $\lambda'-\lambda = \frac{h}{m_ec}(1-\cos\theta).\qquad\blacksquare$ (2)

(c) (3): At $\theta=90°$, $\cos\theta=0$, $\Delta\lambda = h/m_ec = 2.426\times10^{-12}\,\text{m} = 2.43\,\text{pm}$. (2) Since $\Delta\lambda$ is fixed at ~pm scale, the fractional shift $\Delta\lambda/\lambda$ is negligible for visible light ($\lambda\sim500$ nm) but comparable to X-ray wavelengths ($\sim$ 0.1 nm), so measurable there. (1)

Q2 — Particle in a box (12)

(a) (5): Inside, TISE: $-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}=E\psi$ (1). Let $k^2=2mE/\hbar^2$: $\psi''+k^2\psi=0 \Rightarrow \psi=A\sin kx + B\cos kx$ (1). BC $\psi(0)=0\Rightarrow B=0$ (1). BC $\psi(L)=0\Rightarrow \sin kL=0 \Rightarrow kL=n\pi$, $n=1,2,3...$ (2).

(b) (3): Normalise: $\int_0^L A^2\sin^2(n\pi x/L)dx = A^2 L/2 =1 \Rightarrow A=\sqrt{2/L}$ (2). $\psi_n(x)=\sqrt{\frac2L}\sin\!\left(\frac{n\pi x}{L}\right).$ (1)

(c) (4): $k=n\pi/L$ with $E=\hbar^2k^2/2m$ gives $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}.$ (2) Ground state $n=1$, $L=0.20\,\text{nm}$: $E_1 = \frac{(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(0.20\times10^{-9})^2} = 1.506\times10^{-18}\,\text{J} = 9.4\,\text{eV}.$ (2)

Q3 — Time dilation (10)

(a) (6): Postulates (1): (i) laws of physics identical in all inertial frames; (ii) $c$ same in all inertial frames. In clock rest frame: light travels $2d$, so $\Delta t_0 = 2d/c$ (1). In lab frame, mirror moves; light travels along hypotenuses. In time $\Delta t$ the horizontal distance is $v\Delta t$, vertical $2d$. Total path $= 2\sqrt{d^2+(v\Delta t/2)^2}$, and by postulate 2 this $=c\Delta t$ (2). So $c^2\Delta t^2 = 4d^2 + v^2\Delta t^2 \Rightarrow \Delta t^2(c^2-v^2)=4d^2$. With $2d = c\Delta t_0$: $\Delta t = \frac{2d/c}{\sqrt{1-v^2/c^2}} = \gamma\Delta t_0.\qquad\blacksquare$ (2)

(b) (4): $\gamma = 1/\sqrt{1-0.99^2} = 1/\sqrt{0.0199}=7.09$ (2). $\Delta t = 7.09\times2.2\,\mu\text{s} = 15.6\,\mu\text{s}$ (1). Distance $= v\Delta t = 0.99\times3\times10^8\times15.6\times10^{-6} = 4.63\times10^3\,\text{m} \approx 4.6\,\text{km}$ (1).

Q4 — Photoelectric effect (8)

(a) (4): One photon = one quantum $E=hf$ delivered to one electron.

• Threshold: need $hf\geq\phi$; below $f_0=\phi/h$ no single photon has enough energy regardless of intensity. (1.5)
• Instantaneous: energy delivered in a single photon absorption event, not accumulated over time as in wave theory. (1.5)
• KE independent of intensity: intensity = number of photons; each photon's energy sets $KE_{max}=hf-\phi$; more photons → more electrons, not faster ones. (1)

(b) (4): Threshold $\lambda_0 = hc/\phi = \dfrac{1240\,\text{eV·nm}}{2.28\,\text{eV}} = 544\,\text{nm}$ (2). Photon energy at 400 nm: $E=1240/400 = 3.10\,\text{eV}$. $KE_{max}=3.10-2.28 = 0.82\,\text{eV}$ (2).

Q5 — Decay law (9)

(a) (4): $\dfrac{dN}{dt}=-\lambda N$ (1). Separate: $\dfrac{dN}{N}=-\lambda\,dt$, integrate $\ln(N/N_0)=-\lambda t$ (2) $\Rightarrow N=N_0e^{-\lambda t}$. Half-life: $N_0/2=N_0e^{-\lambda t_{1/2}}\Rightarrow t_{1/2}=\ln2/\lambda$ (1).

(b) (5): $\lambda = \ln2/(8.0\,\text{d}) = 0.0866/\text{day} = 1.00\times10^{-6}\,\text{s}^{-1}$ (2). Activity $A=\lambda N \Rightarrow N = A/\lambda = 4.0\times10^{10}/1.00\times10^{-6} = 4.0\times10^{16}$ nuclei (2). After 24 days = 3 half-lives: $A = 4.0\times10^{10}/2^3 = 5.0\times10^9\,\text{Bq}$ (1).

Q6 — Code + relativistic energy (9)

(a) (3):

def rel_energy(m, p, c):
    E = (( p*c )**2 + ( m*c**2 )**2)**0.5   # total energy
    KE = E - m*c**2                          # kinetic energy
    return E, KE

Marks: correct total-energy formula (1), KE = E − rest energy (1), valid structure (1).

(b) (3): $E=\sqrt{2.0^2+0.511^2}=\sqrt{4.0+0.261}=\sqrt{4.261}=2.064\,\text{MeV}$ (2). $KE = 2.064-0.511 = 1.553\,\text{MeV}$ (1).

(c) (3): $E=mc^2\sqrt{1+(pc/mc^2)^2}$ (1). For $pc\ll mc^2$, binomial: $\approx mc^2\left(1+\tfrac12\frac{p^2c^2}{m^2c^4}\right)$ (1) $=mc^2 + \frac{p^2}{2m}$ (1).

[
  {"claim":"Compton shift at 90 deg equals h/m_e c = 2.426 pm","code":"lam_c=2.426e-12; shift=lam_c*(1-0); result=abs(shift-2.426e-12)<1e-15"},
  {"claim":"Particle in box ground energy ~9.4 eV for L=0.20nm","code":"h=6.626e-34; m=9.11e-31; L=0.20e-9; E=h**2/(8*m*L**2); eV=E/1.602e-19; result=abs(eV-9.4)<0.3"},
  {"claim":"Muon lab lifetime 15.6 us and distance ~4.6 km at 0.99c","code":"g=1/(1-0.99**2)**0.5; dt=g*2.2e-6; d=0.99*3e8*dt; result=abs(dt-15.6e-6)<0.3e-6 and abs(d-4630)<200"},
  {"claim":"Photoelectron KE at 400nm with phi=2.28eV is 0.82 eV","code":"E=1240/400; KE=E-2.28; result=abs(KE-0.82)<0.03"},
  {"claim":"Activity after 3 half-lives is 5.0e9 Bq","code":"A=4.0e10/2**3; result=abs(A-5.0e9)<1e8"},
  {"claim":"Relativistic total energy for p=2.0 MeV/c electron is 2.064 MeV, KE 1.553","code":"E=(2.0**2+0.511**2)**0.5; KE=E-0.511; result=abs(E-2.064)<0.01 and abs(KE-1.553)<0.01"}
]