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Structure Of Atom

Question
CBSEENCH11005264

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 higher to lower state. (Planck's constant h = 39. 79 x 10-14 k Js mol-1)

Solution

Here,    Here comma space space space increment straight E space equals space straight E subscript straight n 2 end subscript space minus space straight E subscript straight n 1 end subscript
space space space space space space space space space space space space space space space space space equals space 214.68 space kJ space mol to the power of negative 1 end exponent
straight h space equals space 39.79 space cross times space 10 to the power of negative 14 end exponent kJs space mol to the power of negative 1 end exponent
According to Planck's quantum theory,
                    increment straight E space equals space hv
therefore space space space space space space space space straight v space equals space fraction numerator increment straight E over denominator straight h end fraction                 ....(1)
Substituting the value in (1),
                       straight v space equals space fraction numerator 214.68 space kJ space mol to the power of negative 1 end exponent over denominator 39.79 space cross times space 10 to the power of negative 14 end exponent straight k space Js space mol to the power of negative 1 end exponent end fraction
space space space space equals space 5.39 space cross times space 10 to the power of 14 straight s to the power of negative 1 end exponent