CBSE Class 11-science Answered
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 | 14 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
h
E = 214.68 kJ mol-1, h = 39.79 x 10-14 kJ mol-1
v = 214.68
39.79X10-14
v = 214.68 X 1014
39.79
= 5.39 x 1014 s-1
Answered by Hanisha Vyas | 14 Sep, 2013, 11:46: PM
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