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Q.

An electron is lying initially in the n=4 excited state. The electron de-excites itself to go to n=1 state directly emitting a photon of frequency v41. If the same electron first de-excites to n=3 state by emitting a photon of frequency v43 and then goes from n=3 to n=1 state by emitting a photon of frequency v31, then

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a

v41=v43+v31

b

v41=v43-v31

c

v43=v41+2v31

d

Data Insufficient

answer is A.

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Detailed Solution

According to Ritz Combination Principle

       vmn=vmi+via           for m<i<n

e.g.   v41=v43+v31 or

        v41=v42+v21

Hence, the correct answer is (A).

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