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For elastic scattering by the Yukawa potential, take $\kappa >0$ and use the first Born approximation in the plane-wave normalizationChoose the polar axis along the momentum transfer $\hslash \mathbf{q}$. The angular integral isThe radial volume element cancels the remaining powers of $r$, leavingFor real $A$, the differential scattering cross-section is thereforeThis also has a finite forward limit when $\kappa >0$. The Born amplitude for a Yukawa potential provides a screened route to the Coulomb potential by taking $\kappa \to 0^{+}$ at fixed $q>0$.

For an alpha particle of charge $2e$ scattering from a heavy nucleus of charge $Ze$, set $A=2Z{e}^2/(4\pi {ϵ}_0)$. Neglect recoil, so that the reduced mass is approximately the projectile mass, $E={\hslash }^2{k}^2/(2m)$, and elastic geometry givesThe Coulomb-limit scattering amplitude becomeswhich gives the Rutherford differential cross-sectionThe unscreened Coulomb potential has long range, so taking this limit does not justify the Born approximation for its full scattering state, particularly for a strongly charged target. The exact Rutherford scattering amplitude differs by a phase and has the same modulus. Thus the resulting differential scattering cross-section remains applicable to the ideal nonrelativistic Coulomb problem beyond the weak-potential regime.

To relate the detector counts to the atomic number, first compute the number of target atoms per illuminated area. With molar mass $M=0.197\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the number density integrated through the foil isThe detector's transverse area is already its area normal to the scattered ray. Its small solid angle isThere is no extra $\cos 45^{\circ }$ factor: the foil is traversed normally and the detector area is transverse to its line of sight.

For a thin foil with negligible multiple scattering and energy loss, the detected fraction is $f=\mathcal{N}(d\sigma /d\Omega )\Delta \Omega$. Thus the measurements implyUsing $E=7.68\times 10^{6}e\mathrm{J}=1.23047\times 10^{-12}\mathrm{J}$ in the Rutherford differential cross-section gives

The stated measurements do not reproduce gold's atomic number 79 in this model. The values and units above agree with the original PDF. Substituting $Z=79$ instead would predict $f\simeq 3.17\times 10^{-7}$ with the other inputs unchanged, rather than $3.7\times 10^{-7}$. No stated correction accounts for that discrepancy; forcing the answer to 79 would require changing the data or adding an experimental correction not specified in the problem.

The Zeeman effect comes from $-\mathbf{\mu }\cdot \mathbf{B}$. For an electron, the orbital and spin magnetic moments are approximately $-{\mu }_B\mathbf{L}/\hslash$ and $-2{\mu }_B\mathbf{S}/\hslash$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the spin-orbit interaction: the electron's motion through the atomic electric field produces a magnetic interaction in its rest frame. The relativistic correction, including Thomas precession, and the radial expectation value are incorporated into $K$. This effective Hamiltonian retains the linear Zeeman splitting and spin-orbit coupling, while omitting diamagnetic and hyperfine structure corrections.

Define the angular momentum ladder operators by $L_{\pm }=L_x\pm i{L}_y$ and similarly for $S_{\pm }$. Direct expansion givesso adding $L_z{S}_z$ gives the required scalar product. The off-diagonal terms change $m_{\ell }$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good quantum number at every magnetic field.

In the given product eigenbasis, the diagonal matrix elements areFor $\ell =1$ and $s=1/2$, the only off-diagonal couplings are between $(1,-1/2)$ and $(0,1/2)$, and between $(0,-1/2)$ and $(-1,1/2)$. Each relevant orbital ladder factor is $\hslash \sqrt{2}$ and each spin ladder factor is $\hslash$. Multiplying by $K/{\hslash }^2$ gives $\sqrt{2}K$. ThereforeAll other off-diagonal entries vanish either by the ladder endpoints or by conservation of $m_j$. This derives every entry of the displayed matrix.

The stretched states have no mixing and yieldThe $m_j=1/2$ block is $\left(\begin{array}{cc}-K& \sqrt{2}K\\ \sqrt{2}K& a\end{array}\right)$, while the $m_j=-1/2$ block is $\left(\begin{array}{cc}-a& \sqrt{2}K\\ \sqrt{2}K& -K\end{array}\right)$. Taking the trace and determinant of each $2\times 2$ block gives the remaining four energy eigenvalues:The subscripts indicate $m_j=\pm 1/2$ and the superscripts indicate the upper or lower root. These formulas are the Zeeman spectrum for orbital angular momentum one and remain valid for either sign of $K$.

For the usual ordering $K>0$, including sodium, the zero-field states follow fromThe $j=3/2$ quartet lies at $K$ and the $j=1/2$ doublet at $-2K$, separated by $3K$. In a weak magnetic field, the Landé g-factors giveThese label the curves near $B=0$. In a strong magnetic field, the Paschen-Back effect makes $(m_{\ell },m_s)$ the appropriate approximate labels:for the mixed states, with the stretched-state formulas already exact. The branch connections for $B>0$ areWithin a fixed $m_j$, the nonzero coupling prevents a crossing; this is an avoided crossing. Curves with different $m_j$ may cross because the Hamiltonian does not couple them.

For the sodium estimate, the zero-field frequency separation fixesThe lower $3s$ doublet has shifts $\pm a$. The Sodium D-line Zeeman pattern gives photon-frequency shifts $a/h$ times $\{\pm 2/3,\pm 4/3\}$ for D${}_1$ and $\{\pm 1/3,\pm 1,\pm 5/3\}$ for D${}_2$. Thus the highest-frequency D${}_1$ component and lowest-frequency D${}_2$ component approach each other by $3a/h$. Extrapolating the weak-field Zeeman splitting givesso the crossover occurs at a field of order $10\mathrm{T}$, where $a\sim K$ and the weak-field approximation is becoming marginal.

The exact eigenvalues refine the estimate within this Hamiltonian. The relevant photon energies, with a common zero-field reference omitted, are $E_{+}^{(-)}+a$ and $E_{-}^{(+)}-a$. The corresponding electric dipole transitions have $\Delta m_j=+1$ and $-1$, respectively. Equating them givesIts positive solution is $a/K=\sqrt{10}/3$, hencein the retained model. The crossing is between optical transition frequencies; the participating upper states themselves need not be degenerate.

A quasihole adds one zero at $\eta$ per electron; its local deficit carries positive electric charge $e/m$.^[ref]

Here is a transport counting outline, conditional on an incompressible quantum Hall fluid. Inserting one magnetic flux quantum $\Delta \Phi =h/e$ slowly compared with $\hslash /\Delta$, where $\Delta$ is the excitation energy gap, produces a circulating electric field by Faraday's law. The transverse sheet conductivity sends a radial electric current whose time integral has magnitudeThe opposite electric charge moves to the boundary, so the total electric charge remains conserved. The counted deficit belongs to a collective quasiparticle; no individual electron has been cut into pieces. The argument assumes the gapped fluid and its Hall response, and is not a consequence of the finite two-electron normalization integral.