Question

The wavenumber of the incident radiation in a Raman spectrometer is 20487 cm−1. What is the wavenumber of the scattered Stokes radiation for the J = 2 ← 0 transition of the molecule N2, being the two N atoms the 14 isotope?

Answer 1

The wavenumber of the scattered stokes is 20475.0108cm-1

Data;

• wavelength = 20487 cm-1
• wavenumber = ?
• Δv = ?

Rotational Quantum Number

let's make

J' = excited rotational quantum number

J" = ground rotational quantum number

B = rotational constant

for

J = 2 -- 0

J' = 2, J" = 0

$\delta V = B * 2*3 - B*0(0+1)\\\delta V = 6B$

The rotational constant B for N2 = 1.9982 cm-1

$\delta V = 6 * 1.9982 = 11.9892 cm^-^1$

The wave number of the scattered stokes radiation is given by

$\delta V_s = V_e_x - \delta V$

substituting all values across

$\delta V_s = 20487 - 11.9892\\\delta V_s = 20475.0108 cm^-^1$

The wavenumber of the scattered stokes is 20475.0108cm-1

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