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;