Q.

An electron in 4th excited state of hydrogen atom de excites to first excited state in all possible ways. Number  of Emission spectral lines observed will be

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a

10

b

15

c

6

d

3

answer is C.

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Detailed Solution

Number of spectral lines=nn+12...where n=n2-n1;Fourth excited state=5th orbitFirst excited state=2nd orbit;   n = 5-2 = 3; Number of spectral lines=3×42=6;
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