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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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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