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: Move from wavelength to frequency.
Convert each wavelength to a frequency using $\nu = c/\lambda$. The source frequency is $\nu_0 = c/\lambda_0$ and the Stokes frequency is $\nu_s = c/\lambda_s$. Since $\lambda_s > \lambda_0$, the Stokes photon has lost energy, so $\nu_s < \nu_0$.

Step 2: Get the vibrational frequency of the molecule.
The molecule carries away a fixed frequency $\nu_v = \nu_0 - \nu_s$ from the scattered photon in the Stokes process. This same $\nu_v$ is handed back to the photon in the anti-Stokes process, so the anti-Stokes frequency is $\nu_a = \nu_0 + \nu_v = 2\nu_0 - \nu_s$.

Step 3: Convert back to wavelength.
Since $\nu = c/\lambda$, dividing the frequency relation by $c$ turns it back into the same reciprocal wavelength relation:
\[ \frac{1}{\lambda_a} = \frac{2}{\lambda_0} - \frac{1}{\lambda_s} \]
Using $\lambda_0 = 5000$ Angstrom and $\lambda_s = 5100$ Angstrom:
\[ \frac{1}{\lambda_a} = \frac{2}{5000} - \frac{1}{5100} = 2.0392 \times 10^{-4} \text{ Angstrom}^{-1} \]
\[ \lambda_a = 4903.8 \text{ Angstrom} \]

Final Answer:
Working from frequencies instead of wavenumbers lands on the same anti-Stokes wavelength. \[ \boxed{L = 4904 \text{ Angstrom}} \]