If the energy difference between two electronic sates is 214.68Kj/mol, calculate the frequency of light emitted when an electron drops from the higher to lower states.

Asked by Keshav sultania | 14th Sep, 2013, 12:46: PM

Expert Answer:

Planck's constant = 39.79 x 10-14 kJ mol-1
The frequency (n) of emitted light is related to the energy difference of two levels (DE) as-

?E = hv  or  v?E


 E = 214.68 kJ mol-1, h = 39.79 x 10-14 kJ mol-1


    v =    214.68 



    v =    214.68 X 1014


      = 5.39 x 1014 s-1  

Answered by Hanisha Vyas | 14th Sep, 2013, 11:46: PM

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