Electronic Spectroscopy, NMR and ESR

Solution. A symmetric top has two equal principal moments. Spinning about the symmetry axis \(z\) means the two axes perpendicular to it are equivalent, so \(I_{1}=I_{2}\), while \(I_{3}\) along the figure axis is different.

Solution. By the Boltzmann distribution, including the degeneracies \(g = 2J+1\), \[ \frac{N_{f}}{N_{i}} = \frac{2J_{f}+1}{2J_{i}+1}\, e^{-hc/\lambda k_{B}T} . \] Here the \(D_{2}\) level has \(J=2\) (\(g=5\)) and the \(F_{3}\) level has \(J=3\) (\(g=7\)), so the degeneracy factor is \(5/7\). The exponent is \[ \frac{hc}{\lambda k_{B}T} = \frac{6.626\times10^{-34}\times3\times10^{8}} {3122\times10^{-10}\times1.380\times10^{-23}\times5000} = 9.228 . \] Hence \[ \frac{N_{f}}{N_{i}} = \frac57\,e^{-9.228} = \frac57\times9.83\times10^{-5} = 7.02\times10^{-5} . \]

Solution. For a rigid rotator \(E \propto J(J+1)\) with \(J = 0,1,2,3,\dots\) The ground state is \(J=0\), so the first excited state is \(J=1\) and the fourth excited state is \(J=4\). Therefore \[ \frac{E_{4}}{E_{1}} = \frac{4(4+1)}{1(1+1)} = \frac{20}{2} = 10 \quad\Longrightarrow\quad E_{4} = 10E_{1} = 10\,\mathrm{meV} . \]

Solution. The rotational energy is \(E = \dfrac{\hbar^{2}}{2I}J(J+1)\). Equating the two energies, \[ \frac{J_{A}(J_{A}+1)}{I_{A}} = \frac{J_{B}(J_{B}+1)}{I_{B}} \quad\Longrightarrow\quad \frac{J_{A}(J_{A}+1)}{J_{B}(J_{B}+1)} = \frac{I_{A}}{I_{B}} = 6 . \] Testing the options with \(J_{B}=1\), so that \(J_{B}(J_{B}+1)=2\), we need \(J_{A}(J_{A}+1) = 12\), i.e.\ \(J_{A}=3\). Hence \(J_{A}=3\), \(J_{B}=1\).

Solution. In the far infrared (pure rotational absorption) spectrum the line spacing is \(2B\), so \[ 2B = 20\,\mathrm{cm}^{-1} \quad\Longrightarrow\quad B = 10\,\mathrm{cm}^{-1} . \] The first Stokes line in the rotational Raman spectrum lies at \(6B\) from the exciting line: \[ \Delta\bar\nu = 6B = 6\times10 = 60\,\mathrm{cm}^{-1} . \]

Solution. CH₃OH is a symmetric rotator, \(I_{x}=I_{y}=I \ne I_{z}\). Classically, \[ E = \frac{1}{2I}\left(J_{x}^{2}+J_{y}^{2}\right) + \frac{1}{2I_{z}}J_{z}^{2} . \] Adding and subtracting \(J_{z}^{2}\) so as to build \(J^{2} = J_{x}^{2}+J_{y}^{2}+J_{z}^{2}\), \[ E = \frac{1}{2I}J^{2} + \left(\frac{1}{2I_{z}}-\frac{1}{2I}\right)J_{z}^{2} . \] Quantum mechanically \(J^{2} \to \hbar^{2}l(l+1)\) and \(J_{z}^{2} \to \hbar^{2}m_{l}^{2}\), so \[ E = \frac{\hbar^{2}}{2I}l(l+1) + \frac{\hbar^{2}m_{l}^{2}}{2}\left(\frac{1}{I_{z}}-\frac{1}{I}\right). \]

Solution. For an anharmonic oscillator \(\varepsilon_{v} = \omega_{e}\left(v+\tfrac12\right) - \omega_{e}x_{e}\left(v+\tfrac12\right)^{2}\). The second overtone is \(v = 0 \to v = 3\): \begin{align*} \bar\nu &= \varepsilon_{v=3}-\varepsilon_{v=0}\\ &= \frac72\omega_{e} - \omega_{e}x_{e}\left(\frac72\right)^{2} - \frac{\omega_{e}}{2} + \omega_{e}x_{e}\left(\frac12\right)^{2}\\ &= 3\omega_{e} - \omega_{e}x_{e}\left(\frac{49}{4}-\frac14\right) = 3\omega_{e} - 12\omega_{e}x_{e} = 3\omega_{e}\left(1-4x_{e}\right). \end{align*}

Solution. The harmonic oscillator has equally spaced levels \(E = h\nu\left(n+\tfrac12\right)\) with \(\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{\mred}}\), so the spacing is \(\Delta E = h\nu\) and \[ k = \left(\frac{2\pi}{h}\Delta E\right)^{2}\mred . \] With \(\Delta E = 8.44\times10^{-2}\times1.6\times10^{-19}\ \mathrm{J}\), \[ k = \left(\frac{2\times3.14}{6.626\times10^{-34}} \times 8.44\times10^{-2}\times1.6\times10^{-19}\right)^{2} \times 1.14\times10^{-26} = 186.7\,\mathrm{N}/\mathrm{m} \approx 187\,\mathrm{N}/\mathrm{m}. \]

Solution. The fundamental lies at \(\omega_{e}(1-2x_{e})\) and the first overtone at \(2\omega_{e}(1-3x_{e})\). Their ratio is \[ \frac{2\omega_{e}(1-3x_{e})}{\omega_{e}(1-2x_{e})} = 1.96 \quad\Longrightarrow\quad \frac{1-3x_{e}}{1-2x_{e}} = 0.98 . \] Hence \(1-3x_{e} = 0.98 - 1.96x_{e}\), giving \(0.02 = 1.04x_{e}\) and \(x_{e} = 0.019 \approx 0.02\).

Solution. Infrared \(\leftrightarrow\) vibrational transitions (iii); ultraviolet–visible \(\leftrightarrow\) electronic transitions of valence electrons (i); X-rays \(\leftrightarrow\) inner-shell electrons (iv); \(\gamma\)-rays \(\leftrightarrow\) nuclear transitions (ii).

Solution. Ordering by photon energy: molecular dissociation and ionization sit highest, electronic spectra next, rotational spectra lowest. Hence 1–ii, 2–iii, 3–i.

Solution. The vibrational energy scale is \(E_{V} = \hbar\sqrt{k/M}\) with the spring constant \(k\) fixed by the electronic energy scale, \(k \sim E_{e}/a_{0}^{2}\), while \(E_{e} \sim \hbar^{2}/ma_{0}^{2}\). Then \[ \frac{E_{V}}{E_{e}} = \frac{\hbar\sqrt{k/M}}{\hbar^{2}/ma_{0}^{2}} = \sqrt{\frac{m}{M}} . \] Equivalently: \(E_{V} \propto 1/\sqrt{M}\), and the only dimensionless combination of \(m\) and \(M\) with that dependence is \((m/M)^{1/2}\).

Solution. Take the three activities in turn.

• \(^{1}\)H–\(^{1}\)H: homonuclear, no dipole moment \(\Rightarrow\) infrared inactive.

• \(^{12}\)C–\(^{16}\)O: both nuclei have even \(Z\) and even \(N\), so \(I=0\) \(\Rightarrow\) NMR inactive.

• \(^{1}\)H–\(^{35}\)Cl: heteronuclear (IR active), polarizability changes on rotation (Raman active), and both nuclei have non-zero spin (NMR active). Correct.

• \(^{16}\)O–\(^{16}\)O: homonuclear \(\Rightarrow\) IR inactive, and \(I=0\) \(\Rightarrow\) NMR inactive.

Solution. For \(^{17}\)O₂:

• It is homonuclear, so there is no change in dipole moment during the vibration — infrared inactive.

• The polarizability does change as the molecule rotates — Raman active.

• The \(^{17}\)O nucleus has spin \(I = \tfrac52 \ne 0\) — NMR active.

Hence Raman active and NMR active.

Solution. The Raman displacement is the same on both sides of the exciting line: \[ \Delta\bar\nu = \bar\nu_{AS}-\bar\nu_{0} = \bar\nu_{0}-\bar\nu_{S}, \qquad\text{i.e.}\qquad \frac{1}{\lambda_{AS}}-\frac{1}{\lambda_{0}} = \frac{1}{\lambda_{0}}-\frac{1}{\lambda_{S}} . \] Therefore \[ \frac{1}{\lambda_{AS}} = \frac{2}{\lambda_{0}}-\frac{1}{\lambda_{S}} = \frac{2\lambda_{S}-\lambda_{0}}{\lambda_{0}\lambda_{S}} \quad\Longrightarrow\quad \lambda_{AS} = \frac{\lambda_{0}\lambda_{S}}{2\lambda_{S}-\lambda_{0}} . \] Substituting \(\lambda_{0}=546\,\mathrm{nm}\), \(\lambda_{S}=552\,\mathrm{nm}\): \[ \lambda_{AS} = \frac{546\times552}{2\times552-546}\ \mathrm{nm} = \frac{546\times552}{558}\ \mathrm{nm} = 540.13\,\mathrm{nm} = 540.13\times10^{-7}\ \mathrm{cm}. \] Hence \[ \bar\nu_{AS} = \frac{1}{540.13\times10^{-7}\ \mathrm{cm}} = 18514\,\mathrm{cm}^{-1} . \]

Solution. ESR uses microwaves (iii); NMR uses radio frequency (i); vibrational transitions use the far infrared (iv); electronic transitions use visible radiation (ii). Note that (P-iii) and (Q-i) alone already eliminate three options.

Solution. The first Stokes line of a rotational Raman spectrum is displaced by \(6B\) from the exciting line. Hence \[ 6B = 12.96\,\mathrm{cm}^{-1} \quad\Longrightarrow\quad B = 2.16\,\mathrm{cm}^{-1} . \]

Solution. The spacing between consecutive \(S\)-branch lines of a rotational Raman spectrum is \(4B\): \[ 4B = 243.2\,\mathrm{cm}^{-1} . \] Stokes lines are displaced below the exciting line by \(4B\left(J+\tfrac32\right)\), so moving to the next higher rotational level moves the Stokes line down by one further \(4B\): \[ \bar\nu_{\text{next}} = 17611.4 - 4B = 17611.4 - 243.2 = 17368.2\,\mathrm{cm}^{-1} \approx 17368 . \] (The laser wavelength \(514.5\,\mathrm{nm}\) is not needed — it only fixes the absolute position of the exciting line.)

Solution. From the Boltzmann ratio of the two starting populations, \[ \frac{I_{S}}{I_{AS}} = e^{\,h\nu/k_{B}T} \quad\Longrightarrow\quad T = \frac{hc\bar\nu}{k_{B}\ln\!\left(I_{S}/I_{AS}\right)} . \] With \(\bar\nu = 300\,\mathrm{cm}^{-1} = 3\times 10^{4}\,\mathrm{m}^{-1}\) and \(I_{S}/I_{AS}=4\), \[ T = \frac{6.626\times10^{-34}\times3\times10^{8}\times3\times10^{4}} {1.38\times10^{-23}\times\ln 4} = 311.7\,\mathrm{K} . \] Using the shortcut \(1\,\mathrm{cm}^{-1}\equiv1.44\,\mathrm{K}\) given in the question, \(T = 300\times1.44/\ln4 = 311.6\,\mathrm{K}\). Either way the accepted range is 311 to 312 K.

Solution. With \(r\) the bond length and \(M\) the molecular mass, \[ E_{R} = \frac{\hbar^{2}}{2Mr^{2}}J(J+1), \qquad E_{V} = \sqrt{\frac{k}{M}}\,\hbar\left(v+\tfrac12\right). \] Since \(r \sim a_{0}\) and \(E_{e} \sim \hbar^{2}/ma_{0}^{2}\), \[ E_{R} \sim \frac{\hbar^{2}}{Ma_{0}^{2}} = \frac{m}{M}E_{e}, \] and, with \(k\) fixed by \(E_{e}\) as in Q12, \[ E_{V} \sim \sqrt{\frac{m}{M}}\,E_{e} . \] So (a) and (c) are both correct. Note the resulting hierarchy \(E_{e} : E_{V} : E_{R} = 1 : (m/M)^{1/2} : (m/M)\) — the same \(1 : 10^{-2} : 10^{-4}\) pattern as \(\eqref{eq:energy-scales}\).

Solution. With \(I(X)=0\) the two nuclei are identical bosons, so the total wavefunction must be symmetric under their exchange. The nuclear spin function is necessarily symmetric (there is only one spin state), and the electronic ground state has even parity, so the rotational wavefunction must be symmetric too. Since the rotational wavefunction has parity \((-1)^{J}\), only even \(J\) levels exist. Odd-\(J\) levels are missing altogether, so only even-\(J\) lines appear.

Solution. Visible light (\(\sim2\,\mathrm{eV}\)) can only eject loosely bound electrons — the photoelectric effect (T). X-rays have enough energy for core-level transitions (P), are diffracted by crystal planes (S) and also show the photoelectric effect (T). Gamma rays cause nuclear transitions (Q) and, above \(1.02\,\mathrm{MeV}\), pair production (R). Thermal neutrons have de Broglie wavelengths of order \(1\,\mathrm{Å}\) and are used for crystal structure determination (S).

Solution. The rotational levels of a rigid rotator are \[ E_{J} = \frac{\hbar^{2}J(J+1)}{2I}, \qquad I = \mred r^{2}, \qquad J = 0,1,2,\dots \] For HD the two nuclei have masses \(m_{H}\) and \(2m_{H}\), so \[ \mred = \frac{m_{H}\times 2m_{H}}{m_{H}+2m_{H}} = \frac{2m_{H}}{3} . \] The two lowest levels are \(J=0\) (\(E_{0}=0\)) and \(J=1\) (\(E_{1} = \hbar^{2}/I\)), so \[ \Delta E = \frac{\hbar^{2}}{I} = \frac{\hbar^{2}}{\mred r^{2}} = \frac{3\hbar^{2}}{2m_{H}r^{2}} , \] that is \(\tfrac32\) in the stated units.

Solution. ESR requires unpaired electrons: the technique detects transitions between the Zeeman sublevels of a non-zero electronic spin, so a species with \(S=0\) gives no signal at all. \[ \text{V} = [\text{Ar}]3d^{3}4s^{2},\quad \text{Cr} = [\text{Ar}]3d^{5}4s^{1},\quad \text{Fe} = [\text{Ar}]3d^{6}4s^{2},\quad \text{Zn} = [\text{Ar}]3d^{10}4s^{2}. \] V, Cr and Fe all have partly filled \(3d\) shells and hence unpaired spins. Zinc has a completely filled \(3d\) shell and a filled \(4s\) shell, so \(S=0\) and there is no ESR spectrum.

Solution. The infinite wall at \(x=0\) imposes \(\psi(0)=0\). Of the eigenfunctions of the full harmonic oscillator, only the odd ones vanish at the origin, so the allowed states here are the full oscillator's \(n = 1,3,5,\dots\). Relabelling by \(j = 0,1,2,\dots\) with \(n = 2j+1\), \[ E_{j} = \left(2j+\tfrac32\right)\hbar\omega_{0} + V_{0}. \] Evaluating: \[ E_{0}^{(\mathrm{P})} = \tfrac32(12)\hbar = 18\hbar, \qquad E_{0}^{(\mathrm{Q})} = 18\hbar + 3\hbar = 21\hbar, \] \[ E_{0}^{(\mathrm{R})} = \tfrac32(4)\hbar + 4\hbar = 10\hbar, \qquad E_{1}^{(\mathrm{R})} = \tfrac72(4)\hbar + 4\hbar = 18\hbar, \qquad E_{0}^{(\mathrm{S})} = \tfrac32(14)\hbar = 21\hbar . \] Hence (a) is false (\(18\hbar \neq 21\hbar\)), while (b) \(21\hbar = 21\hbar\), (c) \(18\hbar = 18\hbar\) and (d) \(10\hbar \neq 21\hbar\) are all true.

Solution. Everything is read off the shape of the curves.

(a) True. The equilibrium bond length \(r_{e}\) is the position of the minimum. Q's minimum sits at larger \(r\).

(b) True. Both curves tend to zero as \(r\to\infty\), so \(U(\infty)=0\) is the dissociation limit: a state with total energy \(E<0\) is bound, one with \(E>0\) is not.

(c) True. For small oscillations \(\omega = \sqrt{k/\mred}\) with \(k = \left.\dd^{2}U/\dd r^{2}\right|_{r_{e}}\). The reduced mass is stated to be the same for both, so \(\omega\) depends only on the curvature at the minimum. P's well is markedly narrower, hence more sharply curved, hence \(\omega_{P} > \omega_{Q}\).

(d) False. The dissociation energy is the depth of the well below the \(U=0\) asymptote. P's minimum is the deeper one, so \(D_{e}(\mathrm{P}) > D_{e}(\mathrm{Q})\).

Solution. The first absorption line of a rigid rotator (\(J = 0 \to 1\)) lies at \(2B\). Hence \[ 2B_{1} = 3.842\,\mathrm{cm}^{-1} \Rightarrow B_{1} = 1.921\,\mathrm{cm}^{-1}, \qquad 2B_{2} = 3.673\,\mathrm{cm}^{-1} \Rightarrow B_{2} = 1.8365\,\mathrm{cm}^{-1} . \] Since \(B = h/8\pi^{2}Ic\), \(B \propto 1/I\), so \[ \frac{I_{2}}{I_{1}} = \frac{B_{1}}{B_{2}} = \frac{1.921}{1.8365} = 1.046 . \]

Solution. The rotational levels are \(E = \dfrac{h^{2}}{8\pi^{2}I}J(J+1) = AJ(J+1)\), so the two lowest levels (\(J=0\) and \(J=1\)) are separated by \[ \Delta E = 2A = \frac{\hbar^{2}}{I}, \qquad I = \mred r^{2} . \] For HD, \(\mred = \dfrac{M_{H}M_{D}}{M_{H}+M_{D}} = \dfrac{M_{H}\times2M_{H}}{3M_{H}} = \tfrac23 M_{H}\), so \(\mred c^{2} = \tfrac23\times938\,\mathrm{MeV} = 625.3\,\mathrm{MeV}\). Working in natural units with \(\hbar c = 197.3\,\mathrm{eV}\,\mathrm{nm}\), \[ \Delta E = \frac{(\hbar c)^{2}}{\mred c^{2}\,r^{2}} = \frac{(197.3)^{2}\ \mathrm{eV}^{2}\,\mathrm{nm}^{2}} {625.3\times10^{6}\ \mathrm{eV}\times(0.08)^{2}\ \mathrm{nm}^{2}} = 9.7\times10^{-3}\ \eV \approx 10^{-2}\ \eV . \]

Solution. The lines are equally spaced by \(2B\): \[ 103.73-83.03 = 20.70, \quad 124.30-103.73 = 20.57, \quad 145.03-124.30 = 20.73\ \mathrm{cm}^{-1} . \] Averaging, \[ 2B = \frac{20.70+20.57+20.73}{3} = 20.67\,\mathrm{cm}^{-1} \Rightarrow B = 10.33\,\mathrm{cm}^{-1} = 1033\,\mathrm{m}^{-1} . \] Then, with \(B = h/8\pi^{2}Ic\), \[ I = \frac{h}{8\pi^{2}Bc} = \frac{27.99\times10^{-45}}{1033} = 2.7\times10^{-47}\ \mathrm{kg}\,\mathrm{m}^2 \approx 2.8\times 10^{-47}\,\mathrm{kg}\,\mathrm{m}^2, \] using \(h/8\pi^{2}c = 27.99\times10^{-45}\) in SI units.

Solution. For the anharmonic oscillator \(E_{v} = \left(v+\tfrac12\right)\hbar\omega - x_{e}\left(v+\tfrac12\right)^{2}\hbar\omega\). Levels exist only while the energy still increases with \(v\); the highest level is where \(\dd E_{v}/\dd v = 0\): \[ \hbar\omega - 2x_{e}\left(v_{\max}+\tfrac12\right)\hbar\omega = 0 \quad\Longrightarrow\quad v_{\max} = \frac{1}{2x_{e}}-\frac12 . \] With \(x_{e} = 0.001\), \[ v_{\max} = \frac{1}{0.002}-\frac12 = 500 - 0.5 \approx 500 . \]

Solution. The transition energy is \[ \Delta E = \hbar\omega\,\Delta v + B\left[j'(j'+1)-j(j+1)\right] . \] The largest value needs \(\Delta v = +1\) and the largest allowed increase in the rotational term. With \(j \le 2\) and \(\Delta j = \pm1\), the two possible upward rotational steps are \(j=0\to1\) (giving \(B[2-0]=2B\)) and \(j=1\to2\) (giving \(B[6-2]=4B\)). The larger is \(4B\), so \[ \Delta E_{\max} = \hbar\omega + 4B . \]

Solution. The exciting wavenumber is \[ \bar\nu_{0} = \frac{1}{500\times10^{-7}\ \mathrm{cm}} = 20000\,\mathrm{cm}^{-1} . \] The Raman shift is \(770\,\mathrm{cm}^{-1}\), so \[ \bar\nu_{S} = 20000-770 = 19230\,\mathrm{cm}^{-1} \Rightarrow \lambda_{S} = 520\,\mathrm{nm}, \] \[ \bar\nu_{AS} = 20000+770 = 20770\,\mathrm{cm}^{-1} \Rightarrow \lambda_{AS} = 481.5\,\mathrm{nm}. \] Hence \(481.5\,\mathrm{nm}\) and \(520\,\mathrm{nm}\).

Solution. With \(\Delta J = \pm2\) the Raman shift is \[ \left|\Delta\bar\nu\right| = \varepsilon_{J+2}-\varepsilon_{J} = B(4J+6) . \] The Stokes and anti-Stokes lines therefore sit at \[ \bar\nu_{S} = \bar\nu_{0}-B(4J+6), \qquad \bar\nu_{AS} = \bar\nu_{0}+B(4J+6). \] The closest pair is \(J=0\), and their separation is \[ \bar\nu_{AS}-\bar\nu_{S} = 2B(4J+6)\big|_{J=0} = 12B . \]

Solution. The Rayleigh line is elastic scattering and is always the strongest. Between the two Raman lines, the Stokes process starts from the heavily populated lower level while the anti-Stokes process must start from a sparsely populated upper level, so \[ \frac{I_{S}}{I_{AS}} = e^{\,h\nu_{m}/k_{B}T} > 1 . \] Hence \(I_{R} > I_{S} > I_{AS}\).

Solution. The central line of the shifted band is the pure vibrational (\(Q\)) line, here \(10170\,\mathrm{cm}^{-1}\). The vibrational frequency is the shift from the exciting line: \[ \omega_{0} = 12500-10170 = 2330\,\mathrm{cm}^{-1} . \] The lines at \(10150\), \(10158\) lie on one side and \(10182\), \(10190\) on the other; these are the rotational Raman satellites, spaced by \(4B\): \[ 10158-10150 = 8\,\mathrm{cm}^{-1} = 4B \quad\Longrightarrow\quad B = 2\,\mathrm{cm}^{-1} . \] So the answer is 2330 and 2.

Solution. Rotational dissociation occurs when the rotational energy alone reaches the dissociation energy: \[ E_{J} = hcB\,J(J+1) \ \ge\ D \qquad\Longrightarrow\qquad J(J+1) \ \ge\ \frac{D}{hcB}. \] Convert \(B\) to energy with \(1\,\mathrm{cm}^{-1} \equiv 1.24\times 10^{-4}\,\mathrm{eV}\): \[ hcB = 18.8\times1.24\times10^{-4} = 2.33\times 10^{-3}\,\mathrm{eV}, \] \[ J(J+1) \ \ge\ \frac{4.18}{2.33\times10^{-3}} = 1793 . \] Solving the quadratic \(J^{2}+J-1793 = 0\), \[ J = \frac{-1+\sqrt{1+4\times1793}}{2} = \frac{-1+\sqrt{7173}}{2} = \frac{-1+84.7}{2} = 41.8 , \] so \(J_{\min} = 42\), and the nearest printed option is \(45\).

Solution. In a rotation–vibration band there is no \(Q\) branch, so the two innermost lines straddle the missing band centre \(\tilde\nu_{0}\): \[ R(0) = \tilde\nu_{0} + 2B, \qquad P(1) = \tilde\nu_{0} - 2B . \] Their separation is therefore \(4B\), not \(2B\): \[ 4B = 2906 - 2865 = 41\,\mathrm{cm}^{-1} \qquad\Longrightarrow\qquad B = 10.25\,\mathrm{cm}^{-1} . \] The reduced mass of H³⁵Cl is \[ \mred = \frac{1\times35}{36}\ \mathrm{u} = 0.972\ \mathrm{u} = 1.614\times 10^{-27}\,\mathrm{kg}. \] From \(B = h/(8\pi^{2}c\mred r^{2})\), with \(c\) in \(\mathrm{cm}\,\mathrm{s}^{-1}\) because \(B\) is in \(\mathrm{cm}^{-1}\), \[ r^{2} = \frac{h}{8\pi^{2}c\,\mred B} = \frac{6.626\times10^{-34}} {78.96\times3\times10^{10}\times1.614\times10^{-27}\times10.25} = 1.70\times 10^{-20}\,\mathrm{m}^{2}, \] \[ r = 1.30\times 10^{-10}\,\mathrm{m} = \boxed{1.3\,\mathrm{Å}} . \]

Solution. The rotational energy is \(E = \dfrac{\hbar^{2}}{2I}J(J+1)\) with \(I = \mred R^{2}\): \[ I = 8.35\times10^{-28}\times\left(0.742\times10^{-10}\right)^{2} = 4.597\times 10^{-48}\,\mathrm{kg}\,\mathrm{m}^2 . \] Then \[ \frac{\hbar^{2}}{2I} = \frac{\left(1.05\times10^{-34}\right)^{2}}{2\times4.597\times10^{-48}} = \frac{1.112\times10^{-68}}{9.18\times10^{-48}} = 1.21\times10^{-21}\ \mathrm{J} . \] Converting, \[ \frac{\hbar^{2}}{2I} = \frac{1.21\times10^{-21}}{1.6\times10^{-19}}\ \eV = 7.57\times10^{-3}\ \eV = 7.57\,\mathrm{meV} \approx 7\,\mathrm{meV} . \] Hence \(E_{\text{rot}}(J) \approx 7J(J+1)\) meV.

Solution. For a one-dimensional oscillator the equipartition theorem gives \(\langle K\!.E.\rangle = \tfrac12 k_{B}T\) and \(\langle P\!.E.\rangle = \tfrac12 k_{B}T\), so the total energy is \(\langle E\rangle = k_{B}T\). Equating this to the maximum potential energy \(\tfrac12\beta A^{2}\), \[ k_{B}T = \tfrac12\beta A^{2} \quad\Longrightarrow\quad A = \sqrt{\frac{2k_{B}T}{\beta}} = \sqrt{\frac{2\times1.38\times10^{-23}\times300}{100}} = \sqrt{8.28\times10^{-23}} . \] Hence \(A = 9.09\times10^{-12}\ \mathrm{m} = 0.00909\,\mathrm{nm}\), and as a percentage of the atomic spacing, \[ \frac{A}{a} = \frac{0.00909}{0.256} = 0.0355 = 3.55\,\% . \]

Solution. An ion of charge \(q\) produces a field \(E \propto q/r^{2}\) at the neutral atom. This field induces a dipole moment \(p = \alpha E \propto 1/r^{2}\) in the atom, and the interaction energy of an induced dipole with the field that induced it is \[ V = -\tfrac12\alpha E^{2} \propto -\frac{1}{r^{4}} . \] The attraction is therefore short-ranged compared with a Coulomb interaction. Among the printed options, (a) is the official key.