Zeeman Effect, Paschen–Back Effect and Stark Effect

• P: Stern–Gerlach \(\to\) 3. A beam of silver atoms (\(^{2}S_{1/2}\), \(L=0\)) split into two. With no orbital angular momentum, an even number of components can only come from a half-integer intrinsic angular momentum — electron spin.

• Q: Zeeman effect \(\to\) 4. Splitting into \((2J+1)\) equally spaced sublevels is the direct signature of space quantization.

• R: Franck–Hertz \(\to\) 2. Periodic current dips at fixed voltage intervals demonstrate discrete atomic energy levels.

• S: Davisson–Germer \(\to\) 1. Electron diffraction from nickel demonstrates the wave nature of particles.

Solution. A singlet state means \(S=0\), so this is normal Zeeman with \(g=1\). Inverting \(\eqref{eq:zeeman-wavelength}\), \[ B = \frac{c}{\lambda^{2}}\cdot\frac{4\pi m}{e}\,\Delta\lambda = \frac{3\times10^{8}}{\left(650\times10^{-9}\right)^{2}} \cdot\frac{4\pi\times9.1\times10^{-31}}{1.6\times10^{-19}} \times\left(0.03\times10^{-9}\right) = 1.52\,\mathrm{T} . \]

Solution. \(^{2}P_{3/2}\) has \(m_{J} = \pm\tfrac32, \pm\tfrac12\) (four sublevels); \(^{2}S_{1/2}\) has \(m_{J} = \pm\tfrac12\) (two). Applying \(\Delta m_{J} = 0,\pm1\): \[ \begin{aligned} \Delta m_{J} = 0 &: \ \left(\tfrac12\to\tfrac12\right), \left(-\tfrac12\to-\tfrac12\right) && 2\\ \Delta m_{J} = +1 &: \ \left(\tfrac32\to\tfrac12\right), \left(\tfrac12\to-\tfrac12\right) && 2\\ \Delta m_{J} = -1 &: \ \left(-\tfrac12\to\tfrac12\right), \left(-\tfrac32\to-\tfrac12\right) && 2 \end{aligned} \] giving \(\boxed{6}\) transitions, consistent with Table 4.3: \(\Delta J = 1\), \(J_{<} = \tfrac12\), so \(3(2J_{<}+1) = 6\).

Solution. A singlet-to-singlet transition has \(g_{1}=g_{2}=1\), so the normal Zeeman effect always produces exactly 3 components regardless of \(L\).

Solution. Sodium's \(3\,^{2}D\) has \(J = \tfrac32,\tfrac52\) and \(3\,^{2}P\) has \(J = \tfrac12,\tfrac32\). Applying \(\Delta J = 0,\pm1\) and Table 4.3: \[ \begin{aligned} ^{2}D_{5/2}\to{}^{2}P_{3/2}: & \ \Delta J = 1,\ J_{<}=\tfrac32 \ \Rightarrow\ 3(4) = 12 ,\\ ^{2}D_{3/2}\to{}^{2}P_{3/2}: & \ \Delta J = 0,\ J = \tfrac32 \ \Rightarrow\ 6\left(\tfrac32\right)+1 = 10 ,\\ ^{2}D_{3/2}\to{}^{2}P_{1/2}: & \ \Delta J = 1,\ J_{<}=\tfrac12 \ \Rightarrow\ 3(2) = 6 . \end{aligned} \] (\(^{2}D_{5/2}\to{}^{2}P_{1/2}\) is forbidden: \(\Delta J = 2\).) Total \(= 12+10+6 = \boxed{28}\).

Solution. \(5\,^{1}D_{2}\to5\,^{1}P_{1}\) is singlet–singlet, so normal Zeeman. To resolve we need \(\Delta\lambda \ge 0.01\,\mathrm{nm}\): \[ B = \frac{4\pi m c}{e}\cdot\frac{\Delta\lambda}{\lambda^{2}} = \frac{4\pi\times9.1\times10^{-31}\times3\times10^{8}}{1.6\times10^{-19}} \times\frac{0.01\times10^{-9}}{\left(643.8\times10^{-9}\right)^{2}} = 0.514\,\mathrm{T} \approx 0.52\,\mathrm{T} . \]

Solution. For \(^{2}S_{1/2}\): \(L=0\), \(S=\tfrac12\), \(J=\tfrac12\), so \(g_{J}=2\). The two sublevels \(m_{J}=\pm\tfrac12\) are separated by \[ \Delta E = g_{J}\mu_{\mathrm B}B\left[\tfrac12-\left(-\tfrac12\right)\right] = 2\mu_{\mathrm B}B . \]

Solution. For one outer electron \(j = \ell\pm\tfrac12\); the higher state is \(j = \ell+\tfrac12\), splitting into \[ 2j+1 = 2\ell+2 \ \text{sublevels}. \]

Solution. Strong field with no spin–orbit coupling is Paschen–Back: \(E = \mu_{\mathrm B}B\left(m_{L}+2m_{S}\right)\). For the \(d\)-level, \(L=2\) and \(S=\tfrac12\), giving ten \((m_{L},m_{S})\) pairs:

The values \(+1\), \(0\) and \(-1\) each occur twice; the rest are singly degenerate. So there are \(\boxed{3}\) doubly degenerate states.

Solution. \(^{2}P_{1/2}\to{}^{2}S_{1/2}\) is the sodium \(D_{1}\) line. Here \(\Delta J = 0\) with \(J=\tfrac12\) half-integer, so from Table 4.3 the total is \(6\left(\tfrac12\right)+1 = 4\) components (see also Example 5 and the Formula Summary).

Solution. The transition has \(\Delta m_{j} = +\tfrac12-\left(-\tfrac12\right) = +1\), so it is a \(\sigma^{+}\) component. Viewed along the field, \(\sigma\) components are circularly polarized (Table 4.1).

Solution. \(^{2}D_{3/2}\) has \(m_{J} = \pm\tfrac32,\pm\tfrac12\); \(^{2}P_{1/2}\) has \(m_{J} = \pm\tfrac12\). Since \(\Delta J = 1\) with \(J_{<} = \tfrac12\), Table 4.3 gives \(3(2\cdot\tfrac12+1) = \boxed{6}\) lines.

Solution. From Table 4.1: (a) False — along the field only \(\sigma\) lines appear; (b) True; (c) True; (d) True.

Solution. Include the electron spin. The upper configuration \(n=2\), \(l=1\) gives \(j = \tfrac32,\tfrac12\), i.e.\ the two levels \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\); the lower configuration \(n=1\), \(l=0\) gives only \(^{2}S_{1/2}\). In a weak field this is the anomalous Zeeman effect: each level splits into \(2J+1\) sublevels and the selection rule is \(\Delta m_{J} = 0,\pm1\).

The \(m_{J} = +\tfrac32 \to -\tfrac12\) and \(-\tfrac32 \to +\tfrac12\) pairs are excluded because they need \(\Delta m_{J} = \pm2\). All ten lines are distinct because \(g_{J}\) differs between the levels (\(\tfrac43\), \(\tfrac23\) and \(2\)), which is exactly what makes the pattern anomalous rather than normal.

Solution. The ground state \(\psi_{100}\) has even parity, so \(\lvert\psi\rvert^{2}\) is unchanged under \(\vec r \to -\vec r\) and therefore \(\langle\vec r\,\rangle = 0\). Hence \[ \vec p_{\text{elec}} = -e\,\langle \vec r\,\rangle = 0 . \] More generally, no non-degenerate eigenstate of a parity-symmetric Hamiltonian can carry a permanent electric dipole moment: such a state has definite parity, and \(\vec r\) is an odd operator, so its expectation value vanishes identically.

Solution. The \(3d^{4}\) configuration gives, by Hund's rules, \(S = 4\times\tfrac12 = 2\) and \(L = 2+1+0-1 = 2\). The subshell is less than half filled, so the ground state takes the lowest \(J\) (3.6.3): \[ J = \lvert L-S\rvert = 0 . \] Since \(\vec\mu_{J} = -g_{J}\left(\dfrac{e}{2m}\right)\vec J\) and \(J=0\), \[ \boxed{\mu = 0} . \]

Solution. Only the \(4f^{3}\) electrons matter (\(5s^{2}\) and \(5p^{6}\) are closed). By Hund's rules for \(f^{3}\) (\(\ell = 3\), so \(m_{\ell} = 3,2,1,0,-1,-2,-3\)): \[ S = \tfrac32, \qquad L = 3+2+1 = 6, \qquad J = \lvert L-S\rvert = \tfrac92 \ \ (\text{less than half filled}) . \] The term is \(^{4}I_{9/2}\). With the given \(g_{J} = \tfrac{8}{11}\), \[ \mu_{\text{eff}} = g_{J}\sqrt{J(J+1)}\,\mu_{\mathrm B} = \frac{8}{11}\sqrt{\frac92\cdot\frac{11}{2}}\,\mu_{\mathrm B} = \frac{8}{11}\sqrt{\frac{99}{4}}\,\mu_{\mathrm B} = \boxed{3.62\,\mu_{\mathrm B}} . \]

Solution. From \(\eqref{eq:lande-g}\) with \(J=2\), \(S=1\), \(L=1\): \[ g_{J} = 1 + \frac{2(3) + 1(2) - 1(2)}{2\cdot2\cdot3} = 1 + \frac{6}{12} = \frac{3}{2} . \]

Solution. Normal Zeeman (\(^{1}P\to{}^{1}S\) is singlet–singlet) at \(1\,\mathrm{T}\), \(\lambda = 250\,\mathrm{nm}\): \[ \Delta\lambda = \frac{\lambda^{2}\mu_{\mathrm B}B}{hc} = \frac{\left(250\times10^{-9}\right)^{2}\times9.27\times10^{-24}} {3\times10^{8}\times6.625\times10^{-34}} \approx 3\times 10^{-12}\,\mathrm m = 0.003\,\mathrm{nm} . \]

Solution. \[ \Delta\lambda = \frac{\lambda^{2}}{c}\cdot\frac{eB}{4\pi m} = \frac{\left(660\times10^{-9}\right)^{2}}{3\times10^{8}} \times\frac{1.6\times10^{-19}\times0.3}{4\pi\times9.1\times10^{-31}} = 6.09\times 10^{-12}\,\mathrm m = 6\,\mathrm{pm} . \]

Solution. In an extremely strong field the L–S coupling breaks down and \(J\) is no longer a good quantum number. For \(^{2}P\) (\(L=1\), \(S=\tfrac12\)) the level has \[ (2S+1)(2L+1) = 2\times3 = 6 \ \text{sub-states}, \] but the energy depends only on \(m_{L}+2m_{S}\), and the value \(0\) occurs twice (4.4). Hence \(6-1 = \boxed{5}\) distinct levels.

Solution. Intensities follow the statistical weights \((2J+1)\) of the upper levels: \[ \frac{I(D_{1})}{I(D_{2})} = \frac{2\left(\tfrac12\right)+1}{2\left(\tfrac32\right)+1} = \frac{2}{4} = \frac12 , \] i.e.\ \(1:2\).

Solution. For \(J=1\to J=0\) the lower level does not split, so the three components come from the three \(m_{J}\) values of the upper level, with \(\Delta E = g\mu_{\mathrm B}B\) per step. Converting the wavelength splitting to frequency: \[ \Delta\nu = \frac{c}{\lambda^{2}}\Delta\lambda = \frac{3\times10^{8}}{\left(1849\times10^{-10}\right)^{2}} \times 1.6\times10^{-13} = 0.1404\times 10^{10}\ \mathrm{Hz} . \] Then \[ g = \frac{h\,\Delta\nu}{\mu_{\mathrm B}B} = \frac{6.625\times10^{-34}\times0.1404\times10^{10}} {9.27\times10^{-24}\times0.1} = 1.00 . \]

Solution. A \(p\) state gives \(j = \tfrac12, \tfrac32\), hence terms \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\); an \(s\) state gives \(^{2}S_{1/2}\). At intermediate fields both transitions appear with their full anomalous Zeeman patterns: \[ ^{2}P_{3/2}\to{}^{2}S_{1/2}: 6 \ \text{lines}, \qquad ^{2}P_{1/2}\to{}^{2}S_{1/2}: 4 \ \text{lines}, \] giving \(6+4 = \boxed{10}\).

Solution. Test each candidate lower state against Table 4.3 with upper state \(^{3}D_{2}\) (\(J=2\)): \[ \begin{aligned} ^{3}D_{2}\to{}^{3}F_{2}: & \ \Delta J = 0,\ J=2 \ \Rightarrow\ 6(2) = 12 ,\\ ^{3}D_{2}\to{}^{3}P_{1}: & \ \Delta J = 1,\ J_{<}=1 \ \Rightarrow\ 3(3) = 9 ,\\ ^{3}D_{2}\to{}^{3}P_{2}: & \ \Delta J = 0,\ J=2 \ \Rightarrow\ 6(2) = 12 . \end{aligned} \] \(^{3}F_{1}\) does not exist (\(^{3}F\) has \(J = 2,3,4\) only). Only \(^{3}D_{2}\to{}^{3}P_{1}\) gives nine components.

Solution. With \(S=0\) we have \(J = L = 1\) and \(g_{J} = 1\), so the shift is \[ \Delta\nu = \frac{\mu_{\mathrm B}B}{h}\,g_{J}m_{J} = \left(1.4\,\mathrm{MHz}\,\mathrm{G}^{-1}\right)\times\left(1\,\mathrm{G}\right)\times1\times1 = 1.4\,\mathrm{MHz} . \]

Solution. For a \(p\) orbital, \(L=1\) and \(S=\tfrac12\), so \(J = \tfrac12\) or \(\tfrac32\); the proton has \(I=\tfrac12\). With \(H = \lambda\,\vec I\cdot\vec J\), \[ \Delta E = \frac{\lambda\hbar^{2}}{2} \left[F(F+1)-I(I+1)-J(J+1)\right] . \] Enumerating (in units of \(\lambda\hbar^{2}\)): \[ \begin{aligned} J=\tfrac12,\ F=0 &: \ \tfrac12\left[0-\tfrac34-\tfrac34\right] = -\tfrac34 ,\\ J=\tfrac12,\ F=1 &: \ \tfrac12\left[2-\tfrac34-\tfrac34\right] = +\tfrac14 ,\\ J=\tfrac32,\ F=1 &: \ \tfrac12\left[2-\tfrac34-\tfrac{15}{4}\right] = -\tfrac54 ,\\ J=\tfrac32,\ F=2 &: \ \tfrac12\left[6-\tfrac34-\tfrac{15}{4}\right] = +\tfrac34 . \end{aligned} \] The lowest is \(-\tfrac54\lambda\hbar^{2}\), at \(J=\tfrac32\) and \(\boxed{F=1}\).

Solution. Electric dipole selection rules are \(\Delta\ell = \pm1\) and \(\Delta m_{\ell} = 0,\pm1\). From \(\lvert3,0,0\rangle\) (\(\ell=0\), \(m=0\)) the allowed intermediate states are \[ \lvert2,1,1\rangle, \quad \lvert2,1,0\rangle, \quad \lvert2,1,-1\rangle, \] each equally probable, so each carries probability \(\tfrac13\). Only \(\lvert2,1,0\rangle\) can then decay to \(\lvert1,0,0\rangle\) with \(\Delta m = 0\) — but in fact all three can reach \(\lvert1,0,0\rangle\). The probability of the specified route through \(\lvert2,1,0\rangle\) is therefore \(\tfrac13\).

Solution. Three factors in the given proportionality change on replacing the proton by a positron, that is, on going from hydrogen to positronium (Ps).

The \(g\) factor. A positron has the same magnitude of \(g\) as an electron, \(g \approx 2\), in place of the anomalously large \(g_{p} = 5.59\): this reduces \(\Delta E\) by \(2/5.59\).

The mass. \(m_{p} \to m_{e}\), which raises \(\Delta E\) by \(m_{p}/m_{e} = 1836\).

The radius. The Bohr radius scales as the inverse of the reduced mass. In hydrogen \(\mu \approx m_{e}\); in positronium \(\mu = m_{e}/2\), so \(a_{\text{Ps}} = 2a_{0}\) and \(a^{3}\) grows by a factor \(8\), reducing \(\Delta E\) by \(8\).

Collecting, \[ \frac{\Delta E_{\text{Ps}}}{\Delta E_{\text{H}}} = \frac{2}{5.59}\times 1836 \times \frac{1}{8} = 82.1 , \] and since \(\lambda \propto 1/\Delta E\), \[ \lambda_{\text{Ps}} = \frac{21\,\mathrm{cm}}{82.1} = 0.256\,\mathrm{cm} = \boxed{2.6\,\mathrm{mm}} . \]

Solution. A \(J=1 \to J=0\) line splitting into exactly three components with equal spacing is the normal Zeeman pattern (\(\Delta m_{J} = 0,\pm1\) from the three sublevels of \(J=1\) to the single sublevel of \(J=0\)). The outermost two components differ by \(\Delta m_{J} = 2\), and the wavelength splitting corresponding to an energy splitting \(g_{J}\mu_{B}B\,\Delta m_{J}\) is \[ \delta\lambda = \frac{\lambda^{2}}{hc}\,g_{J}\mu_{B}B\,\Delta m_{J} \qquad\Longrightarrow\qquad g_{J} = \frac{\delta\lambda\, hc}{\lambda^{2}\mu_{B}B\,\Delta m_{J}} . \] Substituting \(\delta\lambda = 3.2\times 10^{-13}\,\mathrm{m}\), \(\lambda = 184.9\times 10^{-9}\,\mathrm{m}\), \(B = 0.1\,\mathrm{T}\), \(\Delta m_{J} = 2\): \[ g_{J} = \frac{3.2\times10^{-13}\times6.626\times10^{-34}\times3\times10^{8}} {(184.9\times10^{-9})^{2}\times9.274\times10^{-24} \times 0.1 \times 2} = \frac{6.36\times10^{-38}}{6.34\times10^{-38}} = 1.00 . \] The magnetic moment of a state of quantum number \(J\) is \[ \mu = g_{J}\sqrt{J(J+1)}\,\mu_{B} = 1.00\times\sqrt{2}\,\mu_{B} = \boxed{1.41\,\mu_{B}} . \]

Solution. The Landé \(g\) factor is \[ g_{j} = 1 + \frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)} . \] With \(j=\tfrac32\), \(l=1\), \(s=\tfrac12\): \[ j(j+1) = \tfrac{15}{4},\qquad s(s+1) = \tfrac34,\qquad l(l+1) = 2, \] \[ g_{j} = 1 + \frac{\tfrac{15}{4}+\tfrac34-2}{2\times\tfrac{15}{4}} = 1 + \frac{\tfrac{10}{4}}{\tfrac{15}{2}} = 1 + \frac{1}{3} = \frac{4}{3}. \] Adjacent magnetic sublevels differ by \(\Delta m_{j} = 1\), so \[ \Delta E = g_{j}\mu_{B}B\,\Delta m_{j} = \boxed{\tfrac43\,\mu_{B}B} . \]