Q65 · CSIR-NET Chemistry, June 2017

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 $(\mathrm{g}(\mathrm{p})=5.6$ and $\mathrm{g}(\mathrm{N})=0.4)$
(a)700 MHz
(b)400 MHz
(c)200 MHz
(d)50 MHz
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
At fixed field ν ∝ g_N: 700 × 0.4/5.6 = 50 MHz.

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