If the energy difference between the electronic states is `214.68 kJmol^(-1)`, calculate the frequency of light emited when an electron drop form the height to the lower state. Planck's constant , `h = 39.79 xx 10^(-14)kJsmol^(-1)`

Answer

`5.395 xx 10^(14) sec^(-1)`

Explanation

`Delta E = hv`, i.e., `(214.68 kJ mol^(-1)) = (39.79 xx 10^(-14)kJsec mol^(-1)) v or v = 5.395 xx 10^(14) s^(-1)`