In an axially symmetric field, NQR transition/s expected for nuclear spin quantum number I = 7/2 is/are :
3
In an axially symmetric electric field gradient (\(\eta = 0\)) the nuclear quadrupole energy levels are \(E_{m}=\frac{e^{2}qQ}{4I(2I-1)}\left[3m^{2}-I(I+1)\right]\).
The energy depends on \(m^{2}\), so the levels \(\pm m\) remain degenerate. For I = 7/2 the distinct levels are \(\pm\frac{1}{2},\ \pm\frac{3}{2},\ \pm\frac{5}{2},\ \pm\frac{7}{2}\), i.e. 4 doubly degenerate levels.
The NQR selection rule is \(\Delta m=\pm 1\), which allows the transitions \(\pm\frac{1}{2}\rightarrow\pm\frac{3}{2}\), \(\pm\frac{3}{2}\rightarrow\pm\frac{5}{2}\) and \(\pm\frac{5}{2}\rightarrow\pm\frac{7}{2}\), giving 3 lines.
Check: the term \(3m^{2}\) changes by 6, 12 and 18 (in units of \(e^{2}qQ/[4I(2I-1)]\)) for these steps, so the 3 frequencies are distinct in the ratio 1 : 2 : 3; in general a half-integer spin I gives I – 1/2 = 3 transitions.
Hence, 3.
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