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The convention used here chooses units in which the numerical values of the reduced Planck constant, electron mass, elementary charge and $4\pi$ times the vacuum permittivity are one.

Band curvature is the Hessian matrix of an energy band as a function of wavevector. Dividing it by ${\hslash }^2$ gives the inverse band effective mass tensor. At a one-dimensional band minimum the curvature is positive, whereas at a maximum it is negative. Near a weak two-wave avoided crossing with reciprocal wavenumber $G$ and coupling magnitude $U$, the two edge curvatures are $E_{\pm }^{\prime \prime }={\hslash }^2/m\pm {\hslash }^4{G}^2/(4m^{2}U)$. This is a crystal response coefficient, not a change in the bare electron mass.

The classical electron radius is the length scale $r_e=e^2/(4\pi {ϵ}_0{m}_e{c}^2)$ formed from the elementary charge, vacuum permittivity, electron mass and speed of light. It is not a measured material radius of an electron. In the classical low-energy scattering model it sets the Thomson cross-section, ${\sigma }_T=8\pi r_e^2/3$.

The 2014 adjustment gives the historical Planck constant, elementary charge and electron mass used for numerical evaluation. These are external reference values, without a claim about the exam handbook's rounding.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

Work in natural units with $c=1$, neglect the electron mass, and put the initial proton at rest with mass $M$. Let $k$ and $k^{\prime }$ be the initial and final electron four-momenta, $q=k-k^{\prime }$, and $Q^2=-q^2>0$. Define the laboratory energy transfer ${\nu }_{\mathrm{lab}}=E-E^{\prime }=P\cdot q/M$. This ${\nu }_{\mathrm{lab}}$ has units of energy; the invariant $P\cdot q$ has units of energy squared. The Bjorken scaling variable and the final hadron invariant mass obeyThese follow by squaring $k-k^{\prime }$ and $P+q$, using $k^2=k^{\prime 2}=0$ and $P=(M,0)$. Solving the Bjorken x definition gives the outgoing energy and then the scattering angle:Alternatively, substituting $Q^2=4E{E}^{\prime }{\sin }^2(\theta /2)$ before solving gives $E^{\prime }=E/[1+2E{\sin }^2(\theta /2)/(Mx)]$. A chosen $x,E,Q^2$ must satisfy $E^{\prime }>0$ and $0\le {\sin }^2(\theta /2)\le 1$. In addition, $W$ must exceed the lightest allowed inelastic hadron threshold; positive $W^2$ alone is insufficient.

For an original numerical example, take $M=0.938\mathrm{GeV}$, $x=0.3$, $E=15\mathrm{GeV}$ and $Q^2=3{\mathrm{GeV}}^2$. Then ${\nu }_{\mathrm{lab}}\simeq 5.330\mathrm{GeV}$, $E^{\prime }\simeq 9.670\mathrm{GeV}$, $\theta \simeq 8.25^{\circ }$, and $W\simeq 2.807\mathrm{GeV}$. Varying $2\le Q^2/{\mathrm{GeV}}^2\le 5$ and $10\le E/\mathrm{GeV}\le 20$ at this same Bjorken x keeps every setting physical and gives $W^2>5.54{\mathrm{GeV}}^2$, above the proton-pion threshold and the stricter chosen continuum selection $W>2\mathrm{GeV}$. These are calculated settings, not historical measurements.

The dimensionless SI volume magnetic susceptibility of number density $n$ of independent spherical bound electrons with mass $m$ and electron-weighted mean-square orbital radius. In a uniform magnetic field, the quadratic minimal-coupling Hamiltonian gives $\Delta E=e^2{B}^2⟨x^2+y^2⟩/(8m)$. Spherical symmetry gives $⟨x^2+y^2⟩=2⟨r^2⟩/3$; differentiating twice with respect to $B$ gives the displayed negative response. Atomic orbitals use the bare electron mass, whereas a shallow donor envelope within an effective mass approximation uses that envelope's effective mass.

Let $m_z={\int }_V{M}_{z}dV$ be the axial magnetic dipole moment, where $M_z$ is the axial magnetization. Write $J_z$ for the internal angular momentum associated with that magnetic dipole moment, and $I\dot{\theta }$ for the body's rotational angular momentum, with moment of inertia $I$. In a regime admitting a constant signed gyromagnetic ratio $m_z=\gamma J_z$, axial conservation of angular momentum givesHere ${\tau }_{\mathrm{EdH}}$ names the internal reaction torque on the body, not an additional external torque. The external term includes the suspension. The model assumes that the changing internal angular momentum is transferred to the measured mechanical motion, with other storage or escape channels negligible on the observation timescale. A changing or non-collinear mixture of orbital angular momentum and spin angular momentum need not admit a single constant $\gamma$.

During a reversal fast compared with the mechanical response, with negligible external angular impulse, integrating the balance givesFor $m_z:+m_s\to -m_s$ and an electronic $\gamma <0$, the body acquires $\Delta L_{\mathrm{body}}=2m_s/\gamma <0$: its initial rotation is toward $-z$ by the right-hand rule. Reversing the reversal reverses the impulse. Over a complete periodic cycle, $\Delta m_z=0$, so the net reaction angular impulse vanishes; a resonant alternating response does not imply steadily accumulating angular momentum.

The orbital angular momentum prediction can be derived without a material model. A charge $q$ traveling around a circle of radius $r$ at angular frequency $\omega$ has loop electric current $q\omega /(2\pi )$ and magnetic dipole moment $m=q\omega r^2/2$. Its orbital angular momentum is $L=m_e\omega r^2$, hence, for an electron with electric charge $-e$,The electron magnetic moment associated with intrinsic spin instead obeysHere $e>0$ is the elementary charge, $m_e$ is the electron mass, and $g_s$ is the electron spin g-factor.^[ref] Thus the same change of magnetic dipole moment corresponds to about half as much transferred angular momentum in the pure-spin model as in the orbital angular momentum model. This factor of two is a relation between magnetic dipole moment and angular momentum, not evidence that an electron is literally a rotating charged sphere. In a material with both contributions, the inferred coefficient is a material-dependent magnetomechanical gyromagnetic ratio.

An optical oscillator strength is a dimensionless weight assigned to an optical resonance. In an independent-resonance Lorentz oscillator model using a reference electron mass $m_e$ and a specified participating number density $n$,Its normalization must be stated: assigning one unit of total strength per participating electron means ${\sum }_j{f}_j=1$. A one-resonance approximation with $f_1=1$ represents the whole chosen electron population, rather than asserting that the first allowed interband transition carries all optical strength. The smallest band gap therefore need not equal the fitted resonance energy.

For a reconstructed electron–Positron pair, four-momentum conservation givesNeglecting the electron mass compared with the daughter energies, and calling the opening angle $\theta$, this becomesAs an original example, choose $E_{+}=2\mathrm{GeV}$, $E_{-}=3\mathrm{GeV}$ and $\theta =60^{\circ }$. Then $M_{ee}{c}^2=\sqrt{6}\mathrm{GeV}\simeq 2.449\mathrm{GeV}$, although the pair's laboratory energy is $5\mathrm{GeV}$. The pair carries net momentum. Neither its laboratory energy nor the original proton beam energy is its invariant mass; unobserved products carry additional four-momentum.

For equal, oppositely directed incoming beams with equal total energy $E_b$, their total momentum vanishes. Thus the incoming center-of-mass energy isStipulated $1.50\mathrm{GeV}$ beams therefore provide $3.00\mathrm{GeV}$ of incoming center-of-mass energy. This identity is exact for the stated equal-beam geometry. Radiation before annihilation can reduce the energy forming the final state; the apparatus also has a distribution of beam energies. A nominal machine setting is therefore not an exact event-by-event final-state invariant mass.

The Lamb shift is a radiative displacement of atomic energy levels beyond the values from the Dirac equation in a Coulomb potential. In hydrogen, the $2S_{1/2}$ and $2P_{1/2}$ levels are degenerate for a point Coulomb potential in the Dirac equation, yet the $2S_{1/2}$ level lies higher by roughly a gigahertz. This is distinct from the fine structure splitting between different $J$ levels and from nuclear-spin hyperfine structure.

Electron self-energy in the quantized electromagnetic field is the dominant contribution to the familiar hydrogen splitting; vacuum polarization contributes too. Renormalization separates the physical electron mass from its divergent free-electron self-energy. A useful leading-log scale for a hydrogenic $nS$ level isnot an exact numerical formula. The enhanced sensitivity of S states follows from their nonzero wavefunction at the nucleus; leading short-distance corrections act like contact perturbations. A precision prediction also includes finite nuclear size and recoil, and must consistently separate those effects from the radiative contribution.^[ref]

The original Lamb-Retherford experiment used a beam of hydrogen atoms, electron-impact excitation into metastable $2S$, and microwave-induced transfer into short-lived $2P$ states. Those atoms decayed before reaching the detector. Metastable atoms surviving to a metal target ejected electrons, so a drop in the measured current signalled a resonance. Varying the microwave excitation and magnetic field established the displacement from the Dirac equation expectation. Microwave quenching converted a small energy difference into a measurable loss of a long-lived population, avoiding optical Doppler widths that obscure the splitting.^[ref]

In a weakly doped n-type semiconductor, the mobile electrons occupy states near the bottom of the conduction band. Their local band curvature is positive, so their band effective mass is positive and approximately constant sufficiently near that minimum. For the usual $\beta <0$ convention, expand about $k=0$:Stronger overlap gives larger $|\beta |$, greater curvature and a smaller mass within this model. Positive $\beta$ simply places the minimum at the zone boundary and again gives a positive edge mass. A negative electron mass near a band maximum instead motivates the positive-mass electron hole description; it is not the usual conduction-edge situation for an n-type semiconductor. No numerical comparison with $m_e$ follows without values of $a$ and $\beta$.

Differentiating the tight binding dispersion produced by the Hückel method givesHenceUnlike a free particle, the band effective mass depends on wavevector and can change sign. At an inflection point $\cos (ka)=0$, the inverse mass vanishes and the curvature-based mass diverges; a quadratic approximation then ceases to describe the dispersion locally. For the usual $\beta <0$ convention, the bottom at $k=0$ has positive mass and the top at $k=\pi /a$ has negative electron mass. The constant $\alpha$ has no effect on curvature.

Two derivatives of the free-electron dispersion giveThe band effective mass becomes the ordinary electron mass when no periodic-potential modification of the quadratic dispersion relation is present.

Substitution in part (a) givesandThe conversion uses $1\mathrm{eV}=1.60218\times 10^{-19}\mathrm{J}$. These are the predictions of the stipulated two-dimensional free electron model, with the vacuum electron mass and one carrier per inferred square cell.

A conventional face-centered cubic cell contains four atoms. With three conduction electrons per atom,Using the bare electron mass,The Drude transport relation is $\rho =m_e/(n{e}^2\tau )$, givingFor a degenerate electron gas, transport samples electrons near the Fermi surface, so use $v_F=\hslash k_F/m_e=2.025\times 10^6$ m/s:This is a transport relaxation time and associated transport mean free path inferred from the specified resistivity. It need not equal a single-particle quantum lifetime when scattering is strongly angle dependent. The free-electron valence count also neglects the detailed band structure of aluminium, as the stipulated model requires. A classical thermal speed is not the appropriate velocity scale for this strongly degenerate metal.

The number of formula units per volume follows from the mass density:Each neutral formula unit contains $31+33=64$ electrons, so ${\rho }_e=64n_f=1.5726\times 10^{30}{\mathrm{m}}^{-3}$. Let $R^2$ be the electron-weighted mean-square orbital radius, averaged over all these electrons. The supplied Larmor-Langevin susceptibility then gives the dimensionless SI volume magnetic susceptibilitywhere $a_0=0.0529177\mathrm{nm}$ is the Bohr radius. An explicit rough atomic-scale assumption $R\simeq a_0$ gives ${\chi }_0\simeq -2.6\times 10^{-5}$.

The quoted bulk data do not determine $R$: a nuclear charge does not specify an orbital radius in a screened many-electron chemical atom. Consequently the numerical estimate just made has atomic-model uncertainty; it is not a deduction with guaranteed 30 percent accuracy. For example, the alternative assumed $R=2a_0$ gives four times the magnitude while leaving every supplied bulk number unchanged. Atomic wavefunctions, an explicit screened-orbital model, or measured atomic magnetic susceptibilities of materials exhibiting diamagnetism would be needed to support that accuracy. Keeping $R$ explicit makes the later comparison independent of an unadvertised radius assumption. The bare electron mass belongs in this atomic formula; the conduction band effective mass is relevant instead to the large donor envelope in part (c).

In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.