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Q.

An electron in 4^th 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

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Number of spectral lines= $\frac{n (n + 1)}{2} . . . where n = n_{2} - n_{1};$

Fourth excited state=5th orbit

First excited state=2nd orbit;

n = 5-2 = 3;

Number of spectral lines= $\frac{3 \times 4}{2} = 6;$

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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