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 / 2

In the Franck-Hertz experiment, electrons emitted from a heated cathode are accelerated through a dilute atomic vapour toward a collector. A small retarding potential before the collector rejects electrons that have lost too much kinetic energy. Below the first excitation threshold, mainly elastic collisions occur and the collector current rises as the accelerating voltage increases. At threshold an inelastic collision can transfer a definite excitation energy to an atom, causing a current drop because the slowed electron cannot overcome the retarding potential.

Raising the voltage further permits acceleration again, and repeated losses produce successive current minima. For mercury their spacing is about $4.9\mathrm{V}$, corresponding to an atomic excitation of about $4.9\mathrm{eV}$ and an associated ultraviolet photon near $254\mathrm{nm}$. Contact potentials shift the absolute voltages, so the spacing is more informative than the first minimum's position alone. Discrete energy losses support discrete atomic energy levels; a continuously variable excitation energy would not produce the same recurring fixed loss scale. The experiment establishes quantized excitations without requiring the classical electron orbit of the Bohr model to be literally correct.

A classical orbiting electron accelerates and therefore radiates according to the Larmor formula. Its mechanical energy becomes more negative, its radius decreases, and the orbit spirals inward rather than remaining a stable atom. The Bohr model replaces this prediction with postulated nonradiating stationary states and quantized orbital angular momentum, together with photon emission or absorption only between states, $h\nu =|E_n-E_m|$.

For a heavy nucleus, $m_{e}vr=n\hslash$. Since the de Broglie wavelength is $h/(m_{e}v)$, this impliesThe electronic wave fits an integer number of wavelengths around the orbit. This is a standing-wave consistency interpretation of the quantization rule, not a classical mechanism that prevents an accelerating charge from radiating. For finite nuclear mass, the two-body relative coordinate uses reduced mass $\mu =m_{e}M/(m_e+M)$; the next derivation retains that correction. In the limit $M\gg m_e$, it reduces to the fixed-nucleus electron description above.

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

The emitted photon carries the decrease in atomic energy, so $hc/\lambda =E_n-E_m$ for $n>m$. Using the preceding energy levels givesThe lower state's energy is more negative, making the right-hand side positive. This is the hydrogenic Rydberg formula, with the finite-nuclear-mass factor included in $R_M$.

At room temperature, $k_{B}T$ is of order $0.026\mathrm{eV}$, far smaller than the electronic excitation gaps of the inferred hydrogenic ion. Ground-state absorption is therefore the natural interpretation. Assign the three lines to $1\to 2$, $1\to 3$, and $1\to 4$. Dividing each reciprocal wavelength by its corresponding factor $1-1/n^2$ givesThe small scatter is consistent with the rounded wavelength data. Since $R_{\infty }\simeq 1.0974\times 10^7{\mathrm{m}}^{-1}$ and $\mu \simeq m_e$,The ion is thus hydrogen-like lithium, ${\mathrm{Li}}^{2+}$. If “effective Rydberg constant” denotes only the reduced-mass constant rather than the charge-inclusive spectral coefficient, its value is $R_M\simeq R_{\mathrm{eff}}/9\simeq 1.097\times 10^7{\mathrm{m}}^{-1}$. The two conventions are explicitly distinguished; the rounded lines do not determine a precise isotope mass.

For Coulomb binding, the virial theorem gives a typical kinetic energy of the order of the binding energy. Taking $p_{\mathrm{typ}}\simeq \sqrt{2m_e|E_1|}$ and using the Heisenberg uncertainty relation estimate $r_{\mathrm{typ}}\sim \hslash /p_{\mathrm{typ}}$ givesThe Bohr model gives $r_1=a_0/Z=a_0/2=2.65\times 10^{-11}\mathrm{m}$ for a heavy helium nucleus, agreeing with the characteristic scale. Equivalently, minimizing the localization estimate $E(r)\sim {\hslash }^2/(2m_e{r}^2)-Z{e}^2/(4\pi {ϵ}_{0}r)$ gives that same scale.

This is an order-of-magnitude radius estimate, not an exact deduction of the cloud's expected value from the inequality alone. Numerical coefficients depend on which spread is used; $\Delta x\Delta p\ge \hslash /2$ does not assert $⟨r⟩=\hslash /p$. For the exact hydrogenic $1s$ state, $⟨r⟩=3a_0/(2Z)=3.97\times 10^{-11}\mathrm{m}$ at $Z=2$, one and a half times the Bohr orbital radius. An electron cloud has a radial distribution, whereas the Bohr model assigns a definite radius.