Past exam of the physics A course of the University of Cambridge / 2012 / Paper 1 / Section A

Subtract the two photoelectric effect equations $K_{\max }=hc/\lambda -\Phi$ to eliminate the work function:Here $e$ converts the energy difference from electronvolts to joules. The precision of the rounded stopping energies justifies about two significant figures, $6.7\times 10^{-34}\mathrm{J}\mathrm{s}$.

Let $\omega =\sqrt{k/m}$, $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. With ${\psi }_0={\pi }^{-1/4}{b}^{-1/2}{e}^{-{\xi }^2/2}$, the next two normalized harmonic-oscillator wavefunctions are ${\psi }_1=\sqrt{2}\xi {\psi }_0$ and ${\psi }_2=(2{\xi }^2-1){\psi }_0/\sqrt{2}$. Their energies are $E_n=(n+1/2)\hslash \omega$, with respectively zero, one and two wavefunction nodes, alternating even and odd parity. The Schrödinger equation givesFor harmonic-oscillator wavefunction inflections, the inflections include both classical turning points and simple wavefunction wavefunction nodes. In dimensionless coordinates the complete sets areThe curvature changes sign at every listed point; none of these states has a wavefunction node coinciding with its turning points.

Resolving a length requires momentum transfer $q$ of order $\hslash /\ell$. Taking a hard relativistic scale $qc\sim 4\mathrm{TeV}$ givesThe full de Broglie wavelength convention gives $hc/E\simeq 3.1\times 10^{-19}\mathrm{m}$, differing by $2\pi$; the intended scale is roughly $10^{-19}\mathrm{m}$, far below a proton radius. Two opposing beams have a larger total collision energy, but proton energy is shared among partons: the actual resolution depends on the hard scattering momentum transfer, not simply on beam energy alone.

For radiative temperature uncertainty after background subtraction, write $P_{\mathrm{back}}=B$, so $P_{\mathrm{total}}\simeq 10B$ and $P\simeq 9B$. The two absolute uncertainties are both $0.1B$. If they are independent standard uncertainties,If the percentages instead denote absolute error bounds, linear worst-case propagation gives 0.56%. With correlation coefficient $\rho$, the standard-uncertainty result is $\sqrt{2(1-\rho )}/360$; subtraction makes positive common errors partly cancel. The independence convention is therefore an assumption, not information supplied by the percentages.

For independent electron arrivals, the shot-noise counting model gives mean transferred charge $Q=I\Delta t=10^{-15}\mathrm{C}$ and count $\bar{N}=Q/e\simeq 6.24\times 10^3$. Poisson statistics implyThis is about 79 electron charges rms. It assumes Poisson shot noise; charge discreteness by itself does not rule out correlations, thermal noise or other apparatus noise.