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Chapter 10: Problem 19

A diatomic molecule for which \(\tilde{v}=4401.2 \mathrm{cm}^{-1}\) and \(\bar{B}=121.3 \mathrm{cm}^{-1}\) is initially in the state \((\nu=1, J=2)\) In a Raman experiment utilizing \(15873.0 \mathrm{cm}^{-1}\) incident radiation, determine the wavenumber of the scattered radiation for (a) the Q-branch Stokes line, (b) the O-branch Stokes line, (c) the Q-branch anti-Stokes line. How will the wavenumber computed in part (a) change if the effects of anharmonicity are included?

Short Answer

Expert verified
The wavenumbers of the scattered radiation for the Q-branch Stokes line, the O-branch Stokes line, and the Q-branch anti-Stokes line are computed using the given values and formulae. If anharmonicity effects were included, the wavenumber shift for the Q-branch Stokes process would decrease because anharmonicity causes the energy levels to get closer for large vibrational quantum numbers.

Step by step solution

01

Calculating for Q-Branch stokes

For Q-Branch Stokes process, the molecule is initially in the state \((ν=1, J=2)\) and transitions down to \((ν=0,J=2)\). Since \(Δν = -1\) and \(ΔJ = 0\), the total wavenumber shift is given by \(Δν_{St} = ν(0) - ν(1) + 2BJ\). Given that \(ν = 4401.2 cm^{-1}\) and \(B=121.3 cm^{-1}\), we substitute this into the above equation and compute to find \(Δν_{St}\).
02

Calculating for O-Branch stokes

For O-Branch Stokes process, the molecule transitions from \((ν=1,J=2)\) to \((ν=0,J=1)\), so \(Δν = -1\) and \(ΔJ = -1\). The total wavenumber shift is given by \(Δν_{O} = ν(0) - ν(1) + 2(B)(J - 1)\). Substitute the given values into the above equation and solve for \(Δν_{O}\).
03

Calculating for Q-Branch anti-stokes

In the Q-Branch anti-Stokes process, the molecule transitions from \((ν=1,J=2)\) to \((ν=2,J=2)\) so that \(Δν = 1\) and \(ΔJ = 0\). The total wavenumber shift is given by \(Δν_{aS} = ν(2) - ν(1) + 2BJ\). Substitute the given values into the above equation and solve for \(Δν_{aS}\).
04

Accounting for the effects of anharmonicity

Anharmonicity causes a molecule's vibrational energy levels to no longer be equidistant. When anharmonicity is considered, for large vibrational quantum numbers, the energy levels are closer than for a harmonic oscillator. Therefore, the effect of anharmonicity on the wavenumber shift for the Q-branch Stokes process will cause the wavenumber shift to be less than without anharmonicity.

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Most popular questions from this chapter

At very high values of the angular momentum \(J\), the rotational wavenumbers of a linear rotor can go through a maximum due to the presence of centrifugal distortion. Find the value of \(J\) for \(\mathrm{HCl}\left(\bar{B}=10.4400 \mathrm{cm}^{-1} \text {and } \bar{D}=\right.\) \(0.0004 \mathrm{cm}^{-1}\) ) where \(\bar{F}\) is a maximum. Hint: You will need to find a root of a cubic equation.

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