Modern Physics

Question 1

(a) Photon energy $E = hc/\lambda = \dfrac{1240\,\text{eV·nm}}{180\,\text{nm}} = 6.89\,\text{eV}$. (1) $K_{max} = E - \phi = 6.89 - 2.28 = 4.61\,\text{eV}$. (2) Why: Einstein's photoelectric equation — each photon gives all its energy to one electron; the work function is the minimum escape energy.

(b) Compton shift: $\Delta\lambda = \dfrac{h}{m_e c}(1-\cos\theta)$. At $90^\circ$, $\Delta\lambda = \lambda_C = 2.43\times10^{-3}\,\text{nm}$. (2) Scattered wavelength $= 180 + 0.00243 = 180.00243\,\text{nm}$. (1) Fractional shift $\Delta\lambda/\lambda \approx 1.35\times10^{-5}$ — negligibly small. For X-rays ($\lambda\sim0.01$–$0.1\,\text{nm}$) the same absolute shift is a much larger fraction, which is why Compton scattering is observable with X-rays, not visible/UV light. (2) Why: $\Delta\lambda$ is independent of incident wavelength; only the fraction depends on $\lambda$.

(c) Photon: $p = h/\lambda$. Electron with de Broglie wavelength $\lambda$: $p = h/\lambda$ — identical. (2) Numerically $p = \dfrac{6.626\times10^{-34}}{180\times10^{-9}} = 3.68\times10^{-27}\,\text{kg·m/s}$ for both. Neither carries more; de Broglie relation makes momenta equal for equal wavelength. (2) Why: Both obey $p=h/\lambda$; the trick is recognising the "same wavelength" removes any difference.

Question 2

(a) $E_n = \dfrac{n^2 h^2}{8m_e L^2}$. Gap: $E_2 - E_1 = \dfrac{(4-1)h^2}{8m_e L^2} = \dfrac{3h^2}{8m_e L^2} = 3.0\,\text{eV}$. (2) $L^2 = \dfrac{3h^2}{8m_e (3.0\,e)} = \dfrac{h^2}{8m_e\,e}$. (1) $= \dfrac{(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(1.602\times10^{-19})} = 3.76\times10^{-19}\,\text{m}^2$. (1) $L = 6.13\times10^{-10}\,\text{m} \approx 0.61\,\text{nm}$. (1)

(b) $\psi_2 = \sqrt{2/L}\sin(2\pi x/L)$; $P = \int_0^{L/4}\frac{2}{L}\sin^2\!\frac{2\pi x}{L}\,dx$. (2) $= \frac{2}{L}\left[\frac{x}{2} - \frac{L}{8\pi}\sin\frac{4\pi x}{L}\right]_0^{L/4}$. (1) At $x=L/4$: $\sin(\pi)=0$, so $P = \frac{2}{L}\cdot\frac{L}{8} = \frac14 = 0.25$. (2)

(c) For the ground state ($n=1$) the probability in $[0,L/4]$ is less than 0.25 (about $0.091$), because the ground-state $|\psi|^2$ peaks at the centre $L/2$ and is small near the walls, so less probability sits in the outer quarter. (2)

Question 3

(a) $\Delta m = 238.05079 - (234.04360 + 4.00260) = 0.00459\,\text{u}$. (2) $Q = 0.00459 \times 931.5 = 4.27\,\text{MeV}$. (2) Why: Q is the mass defect converted to energy; positive Q ⇒ spontaneous decay.

(b) Momentum conservation: $p_\alpha = p_{Th}$. Energy: $Q = K_\alpha + K_{Th}$, with $K = p^2/2m$. $\dfrac{K_\alpha}{K_{Th}} = \dfrac{m_{Th}}{m_\alpha}$, so $K_\alpha = Q\cdot\dfrac{m_{Th}}{m_{Th}+m_\alpha}$. (3) $K_\alpha = 4.27\times\dfrac{234}{238} = 4.20\,\text{MeV}$. (2) Why: The lighter alpha takes most of the kinetic energy since KE splits inversely to mass at equal momentum.

(c) $\lambda = \dfrac{\ln 2}{T_{1/2}}$. $T_{1/2} = 4.5\times10^9\times3.156\times10^7\,\text{s} = 1.42\times10^{17}\,\text{s}$. (1) $\lambda = 0.693/1.42\times10^{17} = 4.88\times10^{-18}\,\text{s}^{-1}$. (2) Activity $A = \lambda N = 4.88\times10^{-18}\times2.0\times10^{22} = 9.76\times10^{4}\,\text{Bq}$. (2)

Question 4

(a) $\gamma = E/mc^2 = 3.0/0.511 = 5.87$. (1) $v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/34.5} = \sqrt{0.971} = 0.985$. (3)

(b) $(pc)^2 = E^2 - (mc^2)^2 = 3.0^2 - 0.511^2 = 9.0 - 0.261 = 8.739$. (2) $pc = 2.956\,\text{MeV} \Rightarrow p = 2.96\,\text{MeV}/c$. (1)

(c) Tube length in lab $L_0 = 10\,\text{m}$ (proper length of tube — tube is at rest in lab). In electron frame it is contracted: $L = L_0/\gamma = 10/5.87 = 1.70\,\text{m}$. (4) Why: The tube is at rest in the lab, so $10$ m is its proper length; the moving electron sees it Lorentz-contracted. (1)

Question 5

(a) $\Delta p \ge \dfrac{\hbar}{2\Delta x} = \dfrac{1.055\times10^{-34}}{2\times10^{-10}} = 5.27\times10^{-25}\,\text{kg·m/s}$. (2) $K \approx \dfrac{(\Delta p)^2}{2m_e} = \dfrac{(5.27\times10^{-25})^2}{2\times9.11\times10^{-31}} = 1.53\times10^{-19}\,\text{J} = 0.95\,\text{eV}$. (2) This is of order 1 eV — comparable to atomic binding/ionisation energies (a few eV), confirming the uncertainty principle sets the quantum energy scale of atoms. (1)

(b) $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right)$, $R = 1.097\times10^7\,\text{m}^{-1}$. $= 1.097\times10^7(0.25 - 0.0625) = 1.097\times10^7\times0.1875 = 2.057\times10^6\,\text{m}^{-1}$. (3) $\lambda = 4.86\times10^{-7}\,\text{m} = 486\,\text{nm}$ — blue-green (H$_\beta$) line. (2)

[
  {"claim":"Q1a photoelectron KE = 4.61 eV","code":"E=1240/180; K=E-2.28; result=abs(K-4.61)<0.05"},
  {"claim":"Q2b probability in [0,L/4] for n=2 equals 1/4","code":"from sympy import symbols,sin,pi,integrate,Rational; x,L=symbols('x L',positive=True); P=integrate(2/L*sin(2*pi*x/L)**2,(x,0,L/4)); result=simplify(P-Rational(1,4))==0"},
  {"claim":"Q3a Q-value = 4.27 MeV","code":"dm=238.05079-(234.04360+4.00260); Q=dm*931.5; result=abs(Q-4.27)<0.05"},
  {"claim":"Q3b alpha KE = 4.20 MeV","code":"Q=4.27; Ka=Q*234/238; result=abs(Ka-4.20)<0.05"},
  {"claim":"Q4a v/c = 0.985","code":"g=3.0/0.511; v=(1-1/g**2)**0.5; result=abs(v-0.985)<0.005"},
  {"claim":"Q5b Balmer 4->2 wavelength = 486 nm","code":"R=1.097e7; inv=R*(1/4-1/16); lam=1/inv*1e9; result=abs(lam-486)<3"}
]