Q128 · CSIR NET Chemistry, December 2011
Paper: CSIR NET December 2011 · Subject: Physical Chemistry · Chapter: Molecular Spectroscopy · Topic: Vibrational · Marks: 2 · Difficulty: Medium
The vibrational energy levels, $v''=0$ and $v'=1$ of a diatomic molecule are separated by $2143\,\mathrm{cm^{-1}}$. Its anharmonicity $\omega_e x_e$ is $14\,\mathrm{cm^{-1}}$. The values of $\omega_e$ (in $\mathrm{cm^{-1}}$) and first overtone ($\mathrm{cm^{-1}}$) of this molecule are respectively.
(a)2143 and 4286
(b)2157 and 4286
(c)2157 and 4314
(d)2171 and 4258
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed A; on checking, D is correct — see the explanation.
Printed A; the fundamental is ω_e − 2ω_eχ_e, not ω_e.
Explanation
Fundamental (v = 0 → 1): ω_e − 2ω_eχ_e = 2143 ⇒ ω_e = 2143 + 2 × 14 = 2171 cm⁻¹. First overtone (v = 0 → 2): 2ω_e − 6ω_eχ_e = 4342 − 84 = 4258 cm⁻¹. Answer: (d) 2171 and 4258.
Study loop for Molecular Spectroscopy