Paper 3 Quantum Physics Answers

Q1 Photoelectrons and Uncertainty

1. $E=hc/\lambda =5.53\times 10^{-19}\mathrm{J}=3.45\mathrm{eV}$. [2]

2. $K_{\max }=3.45-2.10=1.35\mathrm{eV}$ and $V_s=K_{\max }/e=1.35\mathrm{V}$. [3]

3. From $K=p^2/(2m_e)$,

   $$p_{\max }=\sqrt{2m_e{K}_{\max }}=\sqrt{2(9.11\times 10^{-31})(1.35\times 1.60\times 10^{-19})}\approx 6.3\times 10^{-25}\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}.$$

   [3]

4. $\Delta p=0.050(6.3\times 10^{-25})=3.15\times 10^{-26}\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$, hence

   $$\Delta x_{\min }\approx \frac{h}{\Delta p}\approx 2.1\times 10^{-8}\mathrm{m}.$$

   [2]

5. Higher intensity increases the photon arrival rate and hence the number of emitted electrons, but it does not change photon energy, $K_{\max }$ or the specified fractional momentum uncertainty of each electron. Therefore the uncertainty-product estimate is unchanged. [2]

Q2 Uncertainty Principle

1. A narrow wave packet must be made from many wavelengths/wavenumbers. Since momentum is related to wavelength by $p=h/\lambda$, more wavelength spread means more momentum spread. [3]
2. A localised electron, particle passing through a slit, electron in a finite region, or emitted photon wave packet. [1]
3. It follows from wave nature and the mathematical structure of quantum states, not just measurement disturbance or instrument resolution. [2]