Paper: CSIR-NET June 2017 · Subject: Physical Chemistry · Chapter: Molecular Spectroscopy · Topic: Molecular Spectroscopy – General · Marks: 2 · Difficulty: Medium
For a certain magnetic field strength, a free proton spin transition occurs at 700 MHz. Keeping the magnetic field strength constant the ${}^{14}\mathrm{N}$ nucleus will resonate at ($g(p)=5.6$ and $g({}^{14}\mathrm{N})=0.4$)
(a)700 MHz
(b)400 MHz
(c)200 MHz
(d)50 MHz
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
At fixed field the Larmor frequency is proportional to the nuclear g-factor: ν(¹⁴N) = 700 × 0.4/5.6 = 50 MHz.