Past exam of the physics A course of the University of Cambridge / 2019 / Paper 1 / Section A

Write $L=10^{-9}\mathrm{m}$ and use the Heisenberg uncertainty relation separately in the three Cartesian directions. Since a position coordinate is confined to an interval of length $L$, Popoviciu's inequality on variances gives $\Delta x_i\le L/2$. Consequently $\Delta p_i\ge \hslash /L$, and the expectation value of the kinetic energy obeysThus an uncertainty-based lower estimate is $\boxed{K\gtrsim 0.11\mathrm{eV}}$. This confinement energy bound from position support is not sharp: the exact ground-state energy for a cubical infinite square well is $3{\pi }^2{\hslash }^2/(2m_e{L}^2)=1.13\mathrm{eV}$. The Heisenberg uncertainty principle establishes the confinement scale ${\hslash }^2/(m_e{L}^2)$; identifying a box size with an approximate position standard deviation changes its numerical coefficient.

The rotational kinetic energy is $K=L^2/(2I)$. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by ${\hslash }^{2}l(l+1)$, where $l=0,1,2,\dots$. HenceThe three lowest distinct energy levels are thereforeTheir magnetic quantum degeneracy values are $1,3,5$. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

Inside the barrier, the stationary Schrödinger equation gives an exponentially decaying wavefunction with inverse penetration lengthThe amplitude falls approximately as $e^{-\kappa a}$, so the leading exponential estimate for the transmission probability isFor comparison, wavefunction matching at a finite potential step at both edges gives the rectangular-barrier tunnelling probabilityThe difference is the interface prefactor: for an opaque barrier it is $16E(V_0-E)/V_0^2=2.56$. Both give a transmission of order $10^{-6}$; the first is the requested exponential estimate.

Ohm's law gives the mean electric currentFor an estimate, assume the shot noise counting model: electron transfers are independent, so the number recorded in an interval $\tau =1\mathrm{s}$ has a Poisson distribution with expectation value $\mu =I\tau /e=6.2415\times 10^6$ and variance $\mu$. One measurement has relative standard deviation $1/\sqrt{\mu }=4.00\times 10^{-4}$. Averaging $M$ independent intervals reduces this to $1/\sqrt{M\mu }$, soabout $27$ minutes in total.

This is a one-standard deviation, shot noise estimate. A resistance and mean voltage alone do not specify the actual current fluctuations: a real resistor also has Johnson-Nyquist noise, and electron-transfer correlations can change the shot noise. Its temperature, measurement bandwidth and noise model would be needed for an unconditional acquisition time.

For independent small uncertainties interpreted as standard deviations, propagation of uncertainty givesThe angular uncertainty is in radians. Since ${\sigma }_n/n=0.075/1.35=1/18$, the allowed angular contribution to the fractional uncertainty of the numerical aperture is only $\sqrt{0.06^2-(1/18{)}^2}$. ThereforeThe uncertainty in the refractive index already contributes $5.56\%$. If the quoted errors instead mean strict worst-case bounds, first-order absolute errors must be added rather than added in quadrature; that convention would require $|\delta \alpha |\lesssim \tan 60^{\circ }(0.06-1/18)=0.441^{\circ }$. The $2.25^{\circ }$ result uses the usual independent statistical-error convention.