The $v=0$ to 1 vibration-rotation spectrum of a diatomic molecule exhibits transitions for $R(0)$, R1, $P(1)$ and $P(2)$ lines at $2241, 2254, 2216$ and $2203\,\mathrm{cm^{-1}}$, respectively. From this data, we can conclude that the molecule
(a)has rigid rotation and harmonic vibration
(b)has anharmonic vibration
(c)has rotational-vibrational interaction
(d)is affected by nuclear spin-statistics
Answer
Answer: A ✓ checked by 4AB · confidence medium
The source book printed C; on checking, A is correct — see the explanation.
Printed C; the two branches have identical spacing, so B does not change with v.
Explanation
R(1) − R(0) = 13 and P(1) − P(2) = 13 cm⁻¹; solving 4B₁ − 2B₀ = 13 and 4B₀ − 2B₁ = 13 gives B₀ = B₁ = 6.5 cm⁻¹. Equal rotational constants in v = 0 and v = 1 mean no rotation–vibration interaction, and the evenly spaced lines are the rigid-rotor result; nothing in the data shows anharmonicity or spin statistics.