Question 17.7.2: A radiation of wave number 19865 cm^–1 is incident on a samp......
A radiation of wave number 19865 \mathrm{cm}^{–1} is incident on a sample of ^{14}N_2. If the equilibrium bond length is 110 pm, find the wave number of the scattered Stokes radiation for the transition J = 0 → 2.
Step-by-Step
For a homonuclear diatomic molecule the reduced mass
\begin{array}{l}\mu=\frac{14.0067 \times 10^{-3}}{2 \times 6.02214 \times 10^{23}}=1.16293 \times 10^{-26} \mathrm{~kg}\\ \\\text { moment of inertia } \mathrm{I}=\mu r^{2} \\=1.16293 \times 10^{-26} \times\left(110 \times 10^{-12}\right)^{2} \\=1.41 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}\end{array}Rotational constant
B=\frac{h}{8 \pi^{2} I c}=\frac{6.626 \times 10^{-34}}{8 \times(3.14)^{2} \times 1.41 \times 10^{-46} \times 2.998 \times 10^{10}}=1.985 \mathrm{~cm}^{-1}The Stokes line appears towards lower wave number side and for the transition J = 0 → 2 it is at 6B. Therefore, the line will appear at 19865 – 6 × 1.985 = 19853.09 \mathrm{cm}^{–1}
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