Question

Raman spectrum of a molecule was recorded using a source of wavelength 5000 Angstrom. The first Stokes line is observed at 5100 Angstrom. The first anti-Stokes line will appear at a wavelength \(L\) (in Angstrom). The value of \(L\) (rounded off to nearest integer) is

Hint:

Hint:
The anti-Stokes line sits the same wavenumber shift above the source line that the Stokes line sits below it.

Answer 1

Correct Answer: 4904

Step 1: Understanding the Concept:
In Raman scattering, the molecule either takes a quantum of vibrational energy from the incoming light (the Stokes line, longer wavelength, lower energy) or gives up a quantum of vibrational energy to the scattered light (the anti-Stokes line, shorter wavelength, higher energy). Both lines sit the same energy distance from the source line, just on opposite sides.

Step 2: Key Formula or Approach:
Work in wavenumbers (\(1/\lambda\)), since energy shifts add simply there. The vibrational wavenumber shift is:
\[ \Delta\tilde\nu = \frac{1}{\lambda_0} - \frac{1}{\lambda_s} \]
where \(\lambda_0\) is the source wavelength and \(\lambda_s\) is the Stokes line wavelength. The anti-Stokes line sits the same \(\Delta\tilde\nu\) on the other side of \(1/\lambda_0\):
\[ \frac{1}{\lambda_a} = \frac{1}{\lambda_0} + \Delta\tilde\nu = \frac{2}{\lambda_0} - \frac{1}{\lambda_s} \]

Step 3: Detailed Explanation:
With \(\lambda_0 = 5000\) Angstrom and \(\lambda_s = 5100\) Angstrom:
\[ \frac{2}{\lambda_0} = \frac{2}{5000} = 4.000 \times 10^{-4} \text{ Angstrom}^{-1} \]
\[ \frac{1}{\lambda_s} = \frac{1}{5100} = 1.9608 \times 10^{-4} \text{ Angstrom}^{-1} \]
\[ \frac{1}{\lambda_a} = 4.000 \times 10^{-4} - 1.9608 \times 10^{-4} = 2.0392 \times 10^{-4} \text{ Angstrom}^{-1} \]
\[ \lambda_a = \frac{1}{2.0392 \times 10^{-4}} = 4903.8 \text{ Angstrom} \]

Final Answer:
Rounded to the nearest integer, the anti-Stokes line appears at 4904 Angstrom. \[ \boxed{L = 4904 \text{ Angstrom}} \]