A sample was irradiated by 4358 Å line of mercury. A Raman line was observed at 4452 Å.

Calculate (a) Raman shift in $\mathrm{cm}^{–1}$

(b) The wavelength at which antistoke line would appear in the Raman spectrum.

Question

A sample was irradiated by 4358 Å line of mercury. A Raman line was observed at 4452 Å.

Calculate (a) Raman shift in [latex]\mathrm{cm}^{–1}[/latex]

(b) The wavelength at which antistoke line would appear in the Raman spectrum.

Accepted Answer

[latex]\begin{array}{l} ext { (a) } \Delta \bar{\mathrm{v}}_{ ext {Raman }}\left( ext { in cm }^{-1} ight)=\frac{10^{8}}{\lambda_{\operatorname{exc}}\left( ext { in A }^{0} ight)}-\frac{10^{8}}{\lambda_{ ext {Raman }}\left( ext { in A}^{0} ight)}\ \=\frac{10^{8}}{4358}-\frac{10^{8}}{4452}=484.49 \mathrm{~cm}^{-1} \\end{array}[/latex]

(b) The anti stokes line will appear at a frequency 484.49 [latex]\mathrm{cm}^{–1}[/latex] higher than the frequency (in [latex]\mathrm{cm}^{–1}[/latex]) associated with 4358 [latex]A^0[/latex]. Hg line used as source of excitation hence

[latex]\begin{array}{l}\Delta \bar{\mathrm{v}}_{ ext {exc }}\left( ext { in } \mathrm{cm}^{-1} ight)=\frac{10^{8}}{\lambda_{ ext {exc }}\left( ext { in } \mathrm{A}^{0} ight)}=\frac{10^{8}}{4358 \mathrm{~A}^{0}}=2.295 imes 10^{4} \mathrm{~cm}^{-1}\ \ ext { hence } \bar{\mathrm{v}}_{ ext {antistokes }}=2.295 imes 10^{4} \mathrm{~cm}^{-1}+484.49 \mathrm{~cm}^{-1}=2.3434 imes 10^{4} \mathrm{~cm}^{-1} \ \\lambda\left( ext { in A }^{0} ight)=\frac{10^{8}}{\bar{\mathrm{v}}\left(\mathrm{in cm}^{-1} ight)}=\frac{10^{8}}{2.3434 imes 10^{4}}=4267.3 \mathrm{~A}^{0}\end{array}[/latex]

Question 17.7.1: A sample was irradiated by 4358 Å line of mercury. A Raman l......

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