If the energy difference between two electronic states is 214.68 kJ mol^(-1), calculate the frequency of light emitted when an electron drops from the higher to the lower state. Planck's constant, h = 39.79 xx 10^(-14)kJ sec mol^(-1)

If the energy difference between two electronic states is 214.68 kJ mo

1.	If the energy difference between two electronic states is 214.68 kJ mol^(-1), calculate the frequency of light emitted when an electron drops from the higher to the lower state. Planck's constant, h = 39.79 xx 10^(-14)kJ sec mol^(-1)
Answer» SOLUTION :`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)`