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A substance is illuminated with light from a mercury lamp. Two nearest companions with wavelengths of 4244 Å (red companions) and 3885 Å (violet companion) are observed in the combination scattering spectrum. Find the natural frequency of vibration of the molecules of this substance.

Answer»


SOLUTION :The frequency of the nearest red companion in the combination scattering SPECTRUM is KNOWN to be `v_(1)^(red)=v_(0)-Deltaepsi^(VIB)//h`, of the violet companion `v_(1)^(viol)=v_(0)+Deltaepsi_(vib)//h,"where "v_(0)` is the frequency of light. But `Deltaepsi^(vib)=hv`, where v is the natural frequency of VIBRATIONS of a molecule. There fore `v_(1)^(red)=v_(0)-v,v_(1)^("viol")=v_(0)+v`, from which we find the natural frequency of the molecule to be
`v=(v_(1)^(viol)-v_(1)^(red))/2=c/2(1/lamda_(1)^(viol)=1/lamda_(1)^(red))`


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