Question
The rotational Raman spectrum of a diatomic molecule has the first line of a series of Stokes and anti-Stokes lines at , relative of the
of incident light. What would be its bond length.
Answer :
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For a diatomic molecule, the rotational Raman lines are spaced by ( 2B ), where ( B ) is the rotational constant in wavenumbers (( \text{cm}^{-1} )). The first Stokes line corresponds to a transition ( J = 0 \to J = 2 ), giving a shift of ( \Delta \tilde{\nu} = 6B ). Given: [ \Delta \tilde{\nu} = 346 \ \text{cm}^{-1} = 6B ] [ B = \frac{346}{6} = 57.6667 \ \text{cm}^{-1} \approx 57.67 \ \text{cm}^{-1} ] The rotational constant ( B ) is related to the bond length ( r ) by: [ B = \frac{h}{8 \pi^2 c I} \quad \text{with} \quad I = \mu r^2 ] where ( \mu ) is the reduced mass of the molecule. Rewriting: [ r = \sqrt{\frac{h}{8 \pi^2 c \mu B}} ] Assume the diatomic molecule is HCl as an example. Reduced mass: [ \mu = \frac{m_\text{H} m_\text{Cl}}{m_\text{H} + m_\text{Cl}} = \frac{1 \cdot 35}{1 + 35} \ \text{amu} \approx 0.9722 \ \text{amu} ] Convert amu to kg: [ \mu = 0.9722 \times 1.6605 \times 10^{-27} \ \text{kg} \approx 1.614 \times 10^{-27} \ \text{kg} ] Constants: [ h = 6.626 \times 10^{-34} \
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