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For an HF molecule find the number of rotational levels located between the Zeroth and first excited vibrational levels assuming rorational states to be independent of vibrational ones. The natural vibration frequency of this molecule is equal to 7.79.10^(14) rad//s and the distance between the nuclie is 91.7p m. |
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Answer» Solution :If the vibrational frequency is `omega_(0)` the EXCITATION energy of the FIRST vibrational level will be ` ħ omega_(0)`. Thus if there are `J` ratational LEVELS contained in the band between the ground STATE and the first vibrational excitation, then ` ħ omega_(0)=(J(J+1) ħ^(2))/(2I)` where as stated in the problem we have IGNORED any coupling between the two. For `HF` molecule `I=(m_(H)m_(F))/(m_(H)+m_(F))d^(2)= 1.336xx10^(-4) gm cm^(2)` Then `J(J+1)=(2I omega_(0))/( ħ)= 197.4` For `J=14, J(J+1)= 210`. For `J= 13,J(J+1_= 182`. Thus there lie `13` levels between the ground state and the first vibrational excitation. |
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