Past exam of the short options paper of the physics Prelims of the University of Oxford / 2021 / S03 (Quantum ideas)

A blackbody absorbs all incident electromagnetic radiation, at every frequency, direction and polarization: its absorptivity is one. In thermal equilibrium, Kirchhoff's law of thermal radiation makes its emissivity one as well. Its emitted spectrum depends only on its absolute temperature, through Planck's law, rather than on the material of its surface.

A small aperture in a large isothermal cavity approximates a blackbody because radiation entering it undergoes many reflections and is very unlikely to escape before absorption. The equilibrium radiation field is isotropic and unpolarized. The total emitted radiant exitance is $\sigma T^4$ by the Stefan–Boltzmann law; the peak of the wavelength spectrum shifts to shorter wavelengths as temperature increases according to Wien's displacement law. An ideal absorber can therefore glow brightly when hot: “black” describes absorption, not the absence of thermal emission.

Here $h$ is the Planck constant, $\nu$ is the radiation frequency, $c$ is the speed of light in vacuum, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature in kelvin. The quantity $U_{\nu }$ is spectral energy density per unit frequency, so $U_{\nu }d\nu$ is energy per unit volume in the interval $[\nu ,\nu +d\nu ]$, and $h\nu /(k_{B}T)$ is dimensionless.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.

Convert the skin temperature to $T_s=308.15\mathrm{K}$. To locate the peak per unit wavelength in Planck's law, write $x=hc/(\lambda k_{B}T)$ and maximize $x^5/(e^x-1)$. Its positive stationary point obeysThis gives Wien's displacement law, ${\lambda }_{\max }T=2.89777\times 10^{-3}\mathrm{m}\mathrm{K}$. Hencein the thermal infrared. A peak per unit frequency would occur at a different spectral position; this answer uses the conventional wavelength peak.

For an ideal blackbody surface completely surrounded by an isothermal environment, the Stefan–Boltzmann law gives emitted power $A\sigma T_s^4$ and absorbed power $A\sigma T_e^4$. The net outward radiative heat transfer is thereforeThe gross emitted power is about $1022\mathrm{W}$, but about $837\mathrm{W}$ is absorbed back from the surroundings. The difference, rather than the emitted term alone, is the relevant energy loss.

If the temperatures, exposed area, and surroundings remain fixed, the net radiated energy over one day isUsing the supplied conversion givesA dietary Calorie is one kilocalorie, so the result is about 3820 dietary Calories, rather than 3.82 million dietary Calories.

The estimate assumes the entire $2{\mathrm{m}}^2$ surface remains exposed at $35{}^{\circ }\mathrm{C}$ continuously. Clothing reduces direct exposure and presents a cooler outer emitting surface; the skin temperature also varies across the body and decreases in cooler surroundings through changes in blood flow. Body posture and nearby surfaces reduce the effective exchange area, while blankets and warm surrounding surfaces reduce net radiative heat transfer. These effects make the full-area, constant-temperature daily radiative estimate too large for ordinary clothed conditions.

The unit emissivity approximation is already fairly good for skin in the thermal infrared, so reducing emissivity alone is not the main correction. Also, convection, thermal conduction, and evaporation are additional channels of heat transfer; their existence does not subtract from the calculated radiative term. The computed number is a simplified radiative loss, not a complete metabolic energy budget.

In the photoelectric effect, incident light ejects electrons from a material when each absorbed photon supplies sufficient energy to overcome its work function $W$. Conservation of energy gives the maximum emitted kinetic energy,At fixed frequency above threshold, increasing light intensity primarily increases the electron emission rate; it does not increase $K_{\max }$ in the single-photon regime. Emission occurs without the long accumulation delay predicted for weak classical illumination.

Classical electromagnetism distributes incident energy continuously. By itself, that picture suggests that sufficiently intense or sufficiently long illumination could eject electrons at any frequency, and that emitted electron energies should increase with light intensity. It does not account for the observed threshold frequency and the linear dependence of $K_{\max }$ on frequency with slope $h$. Quantized photon absorption accounts for all three features.

The incident photon energy isSubtracting the work function gives $K_{\max }=0.768952\mathrm{eV}=1.2320\times 10^{-19}\mathrm{J}$. This is much less than the electron rest energy, so nonrelativistic kinetic energy is adequate:The magnitude of the retarding stopping potential satisfies $e{V}_s=K_{\max }$, soThe collector must be negative relative to the emitting surface to retard the electrons; the quoted number is the magnitude. Contact-potential corrections are neglected.

The incident photon rate is ${\dot{N}}_{\gamma }=P/(hc/\lambda )\simeq 2.30\times 10^{15}{\mathrm{s}}^{-1}$. The stated quantum efficiency is $\eta =0.005$, so the emitted electron rate is $\eta {\dot{N}}_{\gamma }\simeq 1.15\times 10^{13}{\mathrm{s}}^{-1}$. If all emitted electrons are collected, the photocurrent magnitude isThe light power controls the number of events; the work function already determined the kinetic energy per emitted electron and should not be subtracted from the photon energy when counting incident photons. Incomplete collection would reduce the measured photocurrent below this value.

An electron accelerated from negligible initial kinetic energy through $V$ gains $eV$. The largest possible single photon energy in an X-ray tube is this entire kinetic energy, so the Duane-Hunt law givesThe massive anode can absorb the required recoil momentum with negligible recoil energy; most actual events produce lower-energy bremsstrahlung photons. The acceleration energy $eV$ is valid without approximating the electron speed nonrelativistically.

Work in the initial rest frame of the free electron. If it absorbed a photon of positive energy $E_{\gamma }$, conservation of momentum would give the final electron momentum magnitude $p_f=E_{\gamma }/c$, while conservation of energy would give $E_f=m_e{c}^2+E_{\gamma }$. But the final electron must obey the relativistic energy–momentum relation,Combining these statements giveswhich contradicts $E_{\gamma }>0$. ThusThe initial rest frame always exists for an electron, so this excludes the process in every inertial frame. Compton scattering is allowed because an outgoing photon carries additional energy and momentum. Absorption by an electron bound in matter is also possible because the rest of the system can share recoil.

For a particle on a circle, the tangential speed is $v=r\dot{\varphi }$, the moment of inertia is $I=m{r}^2$, and the angular momentum is $L=I\dot{\varphi }=mrv$. Hence its kinetic energy isThere is no potential energy variation along the circle in the stated model.

Replacing $L$ by the angular momentum operator gives the Hamiltonian operatorThe time-dependent Schrödinger equation is thereforeFor a stationary state, separate variables as $\Psi (\varphi ,t)=\psi (\varphi )T(t)$. Dividing by the product separates a function of $t$ from one of $\varphi$:Thus $T=e^{-iEt/\hslash }$ up to a constant, and the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2I)=E\psi$.

Diagonalize the commuting angular momentum operator at the same time: $-i\hslash {\psi }^{\prime }=\ell \psi$ has solutions $A{e}^{i\ell \varphi /\hslash }$. A single-valued wavefunction requires $e^{2\pi i\ell /\hslash }=1$, hence $\ell =n\hslash$ with $n\in \mathbb{Z}$. Normalizing with the angular probability measure $d\varphi$ givesThese form an orthonormal basis under ${\int }_0^{2\pi }{\psi }_m^{\ast }{\psi }_{n}d\varphi ={\delta }_{mn}$. The periodic boundary conditions apply to both the wavefunction and its first derivative, ensuring the physical self-adjoint Hamiltonian operator. For probability per arc length $ds=rd\varphi$, the normalization instead reads ${\psi }_n=e^{in\varphi }/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are superpositions of these energy eigenstates.

Applying the Hamiltonian operator to ${\psi }_n$ gives ${\psi }_n^{\prime \prime }=-n^2{\psi }_n$, and thereforeThe ground state $n=0$ has zero kinetic energy. Every positive energy level is of quantum degeneracy two, because $n$ and $-n$ correspond to opposite angular momenta with equal squared magnitude.

The confinement scale suggests a tangential momentum scale $\hslash /r$ and thus an excitation energy scale ${\hslash }^2/(m{r}^2)$, consistent with the Heisenberg uncertainty principle. More precisely the de Broglie wavelength must fit an integer number of times around the circumference: $2\pi r=|n|\lambda$, so $|p|=h/\lambda =|n|\hslash /r$. This explains both the $r^{-2}$ dependence and the $n^2$ sequence.

One must not infer a nonzero ground-state energy by applying $\Delta \varphi \Delta L\ge \hslash /2$ naively. Angle is periodic and is not a globally canonical position operator on the circle. The uniform $n=0$ wavefunction, with $\Delta L=0$, is a concrete counterexample to that naive inequality. The ordinary Cartesian Heisenberg uncertainty principle is not violated by the zero-angular momentum state.

The emitted photon energy is the difference of the two energy levels:Using $E_{\gamma }=hc/\lambda$ givesNo numerical mass or radius is supplied, so this is the determinate result. It is the wavelength conditional on the transition occurring. An isolated particle on a circle with no coupling would remain in its energy eigenstate; a charged particle coupled to electromagnetic radiation can emit subject to its transition selection rules. The $3\to 2$ change has $\Delta n=-1$, compatible with an electric-dipole angular factor for a charged ring particle.