InterviewSolution
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InterviewSolution
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The natural vibration frequency of a hydrogen molecule is equal to 8.25.10^(14)S^(-1), the distance between the nuclei is 74 p m, Find the ratio of the number of these molecules at the first excited vibrational level (v=1) to the number of molecules at the first excited rotational level (J=1) at a temperature T= 875K. It should be remembered that degeneracy of rotational levels is equal to 2J+1. |
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Answer» Solution :The energy of the molecule in the first roatational level will be `( ħ^(2))/(I)`. The ratio of the NUMBER of molecule at the first excited vibrational level to the number of molecule at the first excited rataional level is `(e^(- ħ omega//kT))/((2J+1)e^(- ħ^(2)J(J+1)//2IkT))` `(1)/(3)e^(- ħ omega//kT)=(1)/(3)e^(- ħ(omega-2B)//kT)` where `B=ħ//2I` For the hudrogen molecule `I=(1)/(2)m_(H)d^(2)` `=4.58xx10^(-41)gm cm^(2)` Substitution GIVES `3.04xx10^(-4)` |
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