Your Answer : (Not Answered)

Correct Answer : A

The included terms are the kinetic energy of the electrons and the electron-nucleus, electron-electron and nucleus-nucleus Coulomb energies.

In the Born-Oppenheimer approximation the nuclei are so much heavier than the electrons that they are treated as fixed while the electronic problem is solved. The nuclear kinetic energy is therefore dropped from the electronic Hamiltonian, which keeps:

• the kinetic energy of the electrons;
• the electron-nucleus attraction;
• the electron-electron repulsion;
• the nucleus-nucleus repulsion, which is a constant for a given nuclear arrangement.

The options with the kinetic energy of the nuclei instead of the electrons have it the wrong way round, and the option with no kinetic energy at all leaves the electrons without the term that gives them their quantum energy levels.

Your Answer : (Not Answered)

Correct Answer : C

There are 6 permitted transitions.

In a weak field, \({}^{2}P_{3/2}\) splits into four levels (\(M_{J} = \pm 3/2, \pm 1/2\), \(g = 4/3\)) and \({}^{2}S_{1/2}\) into two (\(M_{J} = \pm 1/2\), \(g = 2\)). The selection rule \(\Delta M_{J} = 0, \pm 1\) allows:

• from \(+3/2\): to \(+1/2\) only
• from \(+1/2\): to \(+1/2\) and \(-1/2\)
• from \(-1/2\): to \(+1/2\) and \(-1/2\)
• from \(-3/2\): to \(-1/2\) only

That is 6 transitions, the anomalous Zeeman pattern of the \(D_{2}\) line. 4 is the count for \({}^{2}P_{1/2} \to {}^{2}S_{1/2}\), and 2 or 10 do not fit the selection rule.

Your Answer : (Not Answered)

Correct Answer : D

The ground term is \({}^{5}D_{0}\).

Apply Hund's rules to \(3d^{4}\):

• Maximum S: four parallel spins, \(S = 2\), multiplicity \(2S + 1 = 5\).
• Maximum L for that S: \(m_{l} = 2, 1, 0, -1\), so \(L = 2\), a D term.
• The shell is less than half full, so \(J = |L - S| = 0\).

The term is \({}^{5}D_{0}\). \({}^{1}D_{2}\) has the wrong spin, and the terms with half-integer J require an odd number of electrons.

Your Answer : (Not Answered)

Correct Answer : B

The field is 321.5 mT.

The ESR condition is \(h\nu = g\mu_{B}B\):

\[ \begin{aligned} B &= \frac{h\nu}{g\mu_{B}} \\ &= \frac{6.626 \times 10^{-34} \times 9 \times 10^{9}}{2 \times 9.2732 \times 10^{-24}} \\ &= \frac{5.963 \times 10^{-24}}{1.855 \times 10^{-23}} \approx 0.3215\ \text{T} \end{aligned} \]

That is 321.5 mT, the familiar X-band value near 0.32 T. 600.5 mT and 60.50 mT do not follow, and 400.0 mT would need a frequency of about 11.2 GHz.

Your Answer : (Not Answered)

Correct Answer : A

The resonance frequency is 40 MHz.

The NMR condition is \(h\nu = g_{N}\mu_{N}B\):

\[ \begin{aligned} \nu &= \frac{5.256 \times 5.0504 \times 10^{-27} \times 1.0}{6.626 \times 10^{-34}} \\ &= \frac{2.655 \times 10^{-26}}{6.626 \times 10^{-34}} \approx 4.0 \times 10^{7}\ \text{Hz} \end{aligned} \]

That is 40 MHz. 4.0 MHz and 0.60 MHz are out by powers of ten, and 60 MHz is closer to the proton's 42.6 MHz per tesla but does not follow from these data.

Your Answer : (Not Answered)

Correct Answer : B

The wavelength is about 6.67 Å.

Energy levels of a hydrogen-like atom scale with the reduced mass. For a muon of about 200 electron masses bound to a proton (1836 electron masses), the reduced mass is about \(\frac{200 \times 1836}{2036} \approx 180\) electron masses, so every wavelength is about 180 times shorter than in hydrogen:

\[ \lambda \approx \frac{1216}{180} \approx 6.7 \]

in Ångström units, starting from hydrogen's Lyman-alpha line at 1216 Å.

The longest Lyman-type line is the 2 to 1 transition. The nearest option is 6.67 Å. Ignoring the reduced mass (dividing by 200) gives about 6.1 Å; 13.3, 5.62 and 3.75 Å are much further off.

Your Answer : (Not Answered)

Correct Answer : D

The transitions are (00°1) \(\rightarrow\) (02°0) and (00°1) \(\rightarrow\) (10°0).

The vibrational levels of CO\(_{2}\) are labelled (\(v_{1}v_{2}^{l}v_{3}\)) for the symmetric stretch, the bend and the asymmetric stretch. Nitrogen in the discharge excites the upper laser level, (00°1), one quantum of the asymmetric stretch. From it there are two laser transitions:

• to (10°0), the symmetric stretch: about 10.6 \(\mu\)m, the main line;
• to (02°0), the bending overtone: about 9.6 \(\mu\)m.

Option B pairs the same two transitions with the wavelengths in the opposite order to the question's (9.6 then 10.6). Options A and C describe the relaxation of the lower levels through the bending mode down to the ground state, which empties the lower level but is not the laser emission.

Your Answer : (Not Answered)

Correct Answer : D

The Raman shift is still 606 cm\(^{-1}\).

A Raman shift equals the energy of a molecular vibration (or rotation) expressed in wavenumbers. It is a property of the molecule, not of the exciting light, so changing the source from the mercury line to the He-Ne laser changes the absolute wavelengths of the Raman lines but not their shift from the exciting line.

The other values come from wrongly scaling the shift with the excitation wavelength or frequency. The constant given in the question (printed as n, meaning Planck's constant) is not needed.

Your Answer : (Not Answered)

Correct Answer : B

The rotational constant of \(^{13}C^{16}O\) is B′ = 1.83 cm\(^{-1}\).

The bond length is the same for both isotopes, so \(B = \frac{h}{8\pi^{2}c\mu r^{2}}\) is inversely proportional to the reduced mass \(\mu\):

\[ \begin{aligned} \mu &= \frac{12 \times 16}{28} = 6.857 \\ \mu' &= \frac{13 \times 16}{29} = 7.172 \\ B' &= B\frac{\mu}{\mu'} = 1.92 \times 0.956 \\ &= 1.836\ \text{cm}^{-1} \end{aligned} \]

The heavier isotope has the larger moment of inertia and so the smaller B, which rules out 2.12 cm\(^{-1}\) and 4.24 cm\(^{-1}\). 1.06 cm\(^{-1}\) is far too large a change for one extra neutron. (The value of h is not needed for a ratio.)

Your Answer : (Not Answered)

Correct Answer : D

Auger emission cannot be obtained from hydrogen.

The Auger process needs three electrons: an inner-shell vacancy is created, an electron from a higher level drops into it, and the energy released ejects a third electron (the Auger electron) instead of emerging as an X-ray. An atom must therefore have electrons in at least two shells, at least three in all.

Hydrogen has only one electron (and helium only two), so neither can emit Auger electrons. Carbon, oxygen and nickel have enough electrons; carbon and oxygen Auger lines are in fact standard signals in Auger electron spectroscopy of surfaces.

Your Answer : (Not Answered)

Correct Answer : D

The sequences in increasing order are 3p, 4s, 3d, 4p & 2p, 3s, 3p, 4s.

By the (n + l) rule, orbitals fill in order of increasing n + l, and for equal n + l the one with lower n comes first: 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d and so on. Both sequences in option D follow this order.

Option A's second sequence puts 3d before 4s. Option B puts 3d before 4s in its first sequence. Option C's second sequence puts 5s before 3d and 4f before 4p, both reversed.

Your Answer : (Not Answered)

Correct Answer : C

The Balmer series is \(\frac{1}{\lambda} = \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)R_{H}\), n = 3, 4, 5 ...

The Balmer lines are transitions that end on the n = 2 level of hydrogen. They lie in the visible region: n = 3 to 2 is the red H-alpha line at 656 nm, and the series converges at about 365 nm.

The other options are the other series: lower level 1 is the Lyman series (ultraviolet), lower level 3 the Paschen series and lower level 4 the Brackett series (both infrared).

Your Answer : (Not Answered)

Correct Answer : C

The rules are Δν = ±1 and ΔJ = 0, ±2 respectively.

Raman scattering involves two photons, so it follows the polarisability rather than the dipole moment. For a harmonic vibration the polarisability changes linearly with displacement, giving Δv = ±1 (Stokes and anti-Stokes lines). For rotation of a linear molecule the polarisability repeats twice per turn, giving ΔJ = 0, ±2; ΔJ = 0 is the unshifted Rayleigh line.

ΔJ = 0, ±1 is the rule for infrared rotation-vibration spectra, and including Δv = 0 in the vibrational rule only adds the Rayleigh line, not a vibrational Raman line.

Your Answer : (Not Answered)

Correct Answer : D

The Landé factor is 1.

\[ g = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]

In a singlet state \( S = 0 \), so \( J = L \) and the fraction vanishes: \( g = 1 \). Such states show the normal Zeeman effect, with the splitting of the classical orbital value.

3/2, 5/2 and 1/2 need a non-zero spin; for example the ³P₁ level has g = 3/2. None of them is possible when \( S = 0 \).

Your Answer : (Not Answered)

Correct Answer : C

The number that differs is the number of p electrons in the Cl⁻ ion.

Fe is [Ar] 3d⁶ 4s²; Fe²⁺ loses the two 4s electrons and keeps 3d⁶, so it has 6 d electrons.

• Ne, 1s² 2s² 2p⁶: 6 p electrons.
• Fe atom, 3d⁶: 6 d electrons.
• Mg, 1s² 2s² 2p⁶ 3s²: 6 s electrons.
• Cl⁻, 18 electrons, 1s² 2s² 2p⁶ 3s² 3p⁶: 12 p electrons.

Only Cl⁻ does not match the 6 of Fe²⁺.

Your Answer : (Not Answered)

Correct Answer : C

The factor is \(\frac{4n^{2}}{n^{2} - 4}\), up to a constant.

The fringe width is \(\beta = \lambda D/d\). With the screen distance fixed, keeping \(\beta\) constant requires \(d \propto \lambda\). For the Balmer series,

\[ \begin{aligned} \frac{1}{\lambda} &= R\left(\frac{1}{4} - \frac{1}{n^{2}}\right) = R\,\frac{n^{2} - 4}{4n^{2}} \\ \lambda &= \frac{1}{R} \cdot \frac{4n^{2}}{n^{2} - 4} \end{aligned} \]

so \(d\) must be scaled by \(4n^{2}/(n^{2} - 4)\). The option \((n^{2} - 4)/4n^{2}\) is proportional to \(1/\lambda\), the inverse of what is needed, and the forms with \(n^{2} + 4\) do not come from the Rydberg formula.

Your Answer : (Not Answered)

Correct Answer : D

The ground-state term of oxygen is \({}^{3}P_{2}\).

Oxygen is \(1s^{2}2s^{2}2p^{4}\). Hund’s rules for the four 2p electrons:

1. Maximum spin: place the electrons with \(m_{l} = +1, 0, -1\) spin up and one more spin down, giving \(S = 1\), a triplet.
2. Maximum \(L\) for that spin: the paired electron goes into \(m_{l} = +1\), so \(M_{L} = 1\) and \(L = 1\), a P term.
3. The shell is more than half full, so the largest \(J = L + S = 2\) lies lowest.

\({}^{3}P_{0}\) would be the ground state for a less-than-half-filled p shell such as carbon (\(2p^{2}\)). \({}^{1}D_{2}\) and \({}^{1}S_{0}\) are excited terms of the same configuration, with lower spin.

Your Answer : (Not Answered)

Correct Answer : B

The allowed transition is \({}^{2}\mathrm{P}_{3/2} \rightarrow {}^{2}\mathrm{D}_{5/2}\).

Electric dipole (LS coupling) selection rules: \(\Delta S = 0\), \(\Delta L = 0, \pm 1\) (with a one-electron jump changing \(l\) by \(\pm 1\)), \(\Delta J = 0, \pm 1\), and \(J = 0 \to 0\) forbidden.

For \({}^{2}\mathrm{P}_{3/2} \to {}^{2}\mathrm{D}_{5/2}\): \(\Delta S = 0\), \(\Delta L = +1\), \(\Delta J = +1\). All satisfied.

The others each break a rule. \({}^{1}\mathrm{S}_{0} \to {}^{3}\mathrm{S}_{1}\) changes the spin multiplicity and has \(\Delta L = 0\) between S states. \({}^{2}\mathrm{D}_{5/2} \to {}^{2}\mathrm{P}_{1/2}\) has \(\Delta J = 2\). \({}^{3}\mathrm{P}_{0} \to {}^{5}\mathrm{D}_{0}\) changes the spin and is a \(J = 0 \to 0\) transition.

Your Answer : (Not Answered)

Correct Answer : A

The energy difference is \(\dfrac{3}{2}\) in units of \(\hbar^{2}/(m_{H}r^{2})\).

Rotational levels are E_J = ℏ²J(J + 1)/(2μr²). With m_D = 2m_H, the reduced mass of HD is

\[ \begin{aligned} \mu &= \frac{m_{H} \cdot 2m_{H}}{3m_{H}} = \tfrac{2}{3}m_{H} \\ E_{1} - E_{0} &= \frac{2\hbar^{2}}{2\mu r^{2}} = \frac{3}{2}\,\frac{\hbar^{2}}{m_{H}r^{2}} \end{aligned} \]

2/3 is the reduced mass ratio itself, not the energy. 6 and 4/3 come from using J(J + 1) = 6 for the J = 2 level, or from treating the molecule with the full mass instead of the reduced mass.

Your Answer : (Not Answered)

Correct Answer : C

Only lines of even \(J\) values are present.

The nuclei have spin 0, so they are bosons and the total molecular wavefunction must be symmetric when the two nuclei are exchanged. With nuclear spin 0 the nuclear-spin part is symmetric, and the electronic ground state is even, so the rotational part must be symmetric too. The rotational wavefunction has exchange symmetry \((-1)^{J}\), so only even \(J\) levels exist.

Rotational Raman lines obey \(\Delta J = 0, \pm 2\), which links even levels only to even levels, so every line comes from an even \(J\) and the odd-\(J\) lines are missing altogether, as in the Raman spectrum of CO₂, whose oxygen nuclei also have spin 0.

Lines of all \(J\) would need a heteronuclear molecule. The 3 : 1 intensity alternation belongs to spin-1/2 nuclei such as in H₂. Only odd \(J\) would require the opposite electronic or nuclear symmetry.