Discussion
When excited with mercury line at 435.8 nm, Benzene shows first Raman shift at 606 cm\(^{-1}\). What will be the Raman shift if excited by He-Ne Laser (632.8 nm).
(Given : n = 6.6 × 10\(^{-34}\) Js) :
Answer: D
The Raman shift is still 606 cm\(^{-1}\).
A Raman shift equals the energy of a molecular vibration (or rotation) expressed in wavenumbers. It is a property of the molecule, not of the exciting light, so changing the source from the mercury line to the He-Ne laser changes the absolute wavelengths of the Raman lines but not their shift from the exciting line.
The other values come from wrongly scaling the shift with the excitation wavelength or frequency. The constant given in the question (printed as n, meaning Planck's constant) is not needed.
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