An electron jumps from the fourth orbit to the second orbit of hydrogen atom. Given: the Rydberg's constant R=105cm−1. The frequency, in Hz, of the emittedradiation will be
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a
316×105
b
36×1015
c
916×105
d
916×1015
answer is D.
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Detailed Solution
1λ=R122−142 or fc=R14−116 or f=cR14−116=3×108×107×316 =916×1015Hz