InterviewSolution
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InterviewSolution
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Find the ratio of energies required to excite a diatomic molecule to the first vibrational and to the first rotational level. Calculate that ratio for the following molecules: Molecule, `omega, 10^(14s^(-1)` d, pm (a) `H_(2) 8.3 74` (b) `HI 4.35 160` (c ) `I_(2) 040 267` Here `omega` is the natural vibration frequency of a molecule, `d` is the distance between nuclei. |
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Answer» For the first rotational level `E_(rot)= 2(ħ^(2))/(2I)=(ħ^(2))/(I)` and for the first vibrational level `E_(vib)= .ħ omega` Thus `xi=(E_(vib))/(E_(rof))=(Iomega)/(ħ)` Here `omega=` frequency of vibration. Now `I= mud^(2)=(m_(1)m_(2))/(m_(1)+m_(2))d^(2)` (a) For `H_(2)` moleculeI `=4.58xx10^(-41)gm cm^(2)` and `xi= 36` (b) For HI moleculeI `=4.2xx47xx10^(-40) gm cm^(2)` and `xi= 175` (c ) For `I_(2)` moleculeI `=7.57xx10^(-38) gm cm^(2)` and `xi= 2872` |
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