Question

The fundamental vibrational mode for H2 is at 4159/cm. What is the wavelength of the first anti-Stokes transition if the sample is irradiated at a wavelength of 1064nm?

Answer 1

Fundamental vibrational mode for H2 is at, $\nu$ = 4159 / cm

excitation wavlength, $\lambda$1 = 1064 x 10^-7 cm

wavelength of the first anti-Stokes transition = $\lambda$2

From Raman spectroscopy.

$\nu$ = 1 / $\lambda$1 - 1 / $\lambda$2

4159 = (1/1064 - 1/$\lambda$2) * 10^-7

4.159x10^-4 = 9.398x10^-4 - 1 / $\lambda$2

1 / $\lambda$2 = 5.239x10^-4

$\lambda$2 = 1908.76 nm

So,  the wavelength of the first anti-Stokes transition =  1908.58 nm