Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S3 / 1

Black-body radiation and the photoelectric effect exposed failures of classical continuous-wave energy exchange. Classical equipartition predicted an unbounded short-wavelength thermal spectrum, the ultraviolet catastrophe. Quantized oscillator energies $E_n=nh\nu$ give the mean excitation energy $h\nu /(e^{h\nu /(k_{B}T)}-1)$ and suppress high-frequency modes, producing Planck's law. In the modern description, thermal radiation consists of photons with energy $h\nu$ and Bose occupation numbers.

For the photoelectric effect, increasing intensity below a threshold frequency does not eject electrons, whereas the maximum electron kinetic energy grows with frequency above threshold. Absorption of one photon transfers $h\nu$ to one electron, giving $\boxed{K_{\max }=h\nu -W}$ and threshold ${\nu }_0=W/h$, where $W$ is the work function. Intensity primarily changes the photon arrival rate and hence the photocurrent in the ordinary single-photon regime. Thus changing energy per quantum and changing the number of quanta have distinct consequences. Historically the thermal derivation first quantized matter's energy exchange; it should not be described as an already complete derivation of modern photon theory.

For a photon, $p_{\gamma }=E/c$ and ${\lambda }_{\gamma }=hc/E$. The neutron is nonrelativistic here because $18.8\mathrm{keV}\ll m_n{c}^2$, so $p_n=\sqrt{2m_{n}E}$ and ${\lambda }_n=h/p_n$. Using the supplied common energy givesEqual kinetic energies do not imply equal momenta: the massive particle obeys $E=p^2/(2m_n)$, while the massless photon obeys $E=pc$. The neutron momentum is about $316$ times larger, so its de Broglie wavelength is about $316$ times smaller.

To give the neutron the photon wavelength, set its momentum to $E/c$, not its kinetic energy to $E$ again. ThenIts kinetic energy would then be about $0.188\mathrm{eV}$, connecting this comparison with the next calculation.

For a transverse periodic row of scatterers of spacing $d$, normal incidence makes their incident phases equal. Scattering at angle $\theta$ from that normal creates an outgoing path difference $d\sin \theta$ between neighbours. Constructive interference requires a phase difference $2\pi n$, givingUsing this printed relation as the intended transverse diffraction grating model, the nonrelativistic neutron wavelength at $0.188\mathrm{eV}$ is $\lambda =6.5964\times 10^{-11}\mathrm{m}$. Hence the first-order data giveGeometry qualification: this is not a derivation from reflection at crystal planes parallel to the surface, as the source prose says. The ordinary Bragg's law for planes of normal separation $d$ is $2d\sin {\theta }_B=n\lambda$, where ${\theta }_B$ is the incident glancing angle to those planes. At normal incidence on parallel planes, ${\theta }_B=\pi /2$ and specular reflected rays leave along the normal, so their angle from the normal is zero; they cannot obey the printed positive-order relation $d\sin 0=n\lambda$ for $\lambda >0$. This is a concrete counterexample to identifying the two geometries. The numerical spacing above is therefore conditional on the printed grating-like relation, rather than a forced assertion that the stated Bragg geometry proves it.

For nonrelativistic neutrons, $\lambda \propto E_K^{-1/2}$, so first-order uncertainty propagation gives $|\Delta \lambda |/\lambda =\frac{1}{2}|\Delta E_K|/E_K=0.05$. ThusAt fixed $d$, differentiating the stipulated first-order relation $d\sin \theta =\lambda$ gives $\Delta \theta =\tan \theta \Delta \lambda /\lambda$. ConsequentlyFor photons, $\lambda \propto E^{-1}$, so the same fractional energy uncertainty gives $\Delta {\lambda }_{\gamma }/{\lambda }_{\gamma }=0.10$. Their $18.8\mathrm{keV}$ wavelength almost equals the low-energy neutron wavelength, so the conditional central angle is $0.116973\mathrm{rad}$ andabout twice the neutron angular spread. This difference comes from the different energy-momentum relations, not from substantially different central wavelengths. The small-angle approximation is adequate at $0.117\mathrm{rad}$: replacing $\sin \theta$ by $\theta$ changes it by about ${\theta }^2/6=0.23\%$, and replacing $\tan \theta$ by $\theta$ changes the sensitivity by about $0.46\%$. The numerical angular results retain the preceding geometry qualification; a different physical scattering geometry would require its own angle convention.