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Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 in the He spectrum?


A

n = 4 to n = 2

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B

n = 3 to n = 2

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C

n = 3 to n = 1

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D

n = 2 to n = 1

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Solution

The correct option is D

n = 2 to n = 1


We know that
Wave number ¯ν=1λ=[1n211n22]Z2×R
Therefore,
¯ν=1λ=(122142)RZ2=34R

In H spectrum for the same ¯ν=λ as Z=1, n=1, n2=2


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