In an axially symmetric field, the first NQR transition energy for I = $\frac{3}{2}$ is :
\(+\frac{e^{2}Qq}{2}\)
Nuclear quadrupole resonance arises because a nucleus with spin \(I \ge 1\) has a non-spherical charge distribution, described by the quadrupole moment \(Q\), which interacts with the electric field gradient \(q\) set up by the surrounding electrons.
For an axially symmetric field the asymmetry parameter is zero, and the quadrupole energy levels depend only on the magnitude of the magnetic quantum number, \(|m_I|\). The standard expression is
\(E_{Q} = \frac{e^{2}Qq\,[\,3m_I^{2} - I(I+1)\,]}{4I(2I-1)}\).
For \(I = 2\) the denominator is \(4 \times 2 \times 3 = 24\) and \(I(I+1) = 6\), so the levels are obtained by substituting \(|m_I| = 0, 1, 2\). Because the energy depends on \(m_I^{2}\), states of equal \(|m_I|\) stay degenerate, leaving three distinct levels.
The selection rule for NQR is \(\Delta m_I = \pm 1\), so the first transition is the one between the two lowest-lying of these levels, and taking the difference of the corresponding energies leaves a simple fraction of \(e^{2}Qq\).
Per the official final answer key the answer is option (C).
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