Number of spectral lines obtained in $H e^{+}$ spectra when an electron makes transition from fifth excited state to first excited state will be…..

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answer is 10.

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Detailed Solution

5^th excited state : $n_{2} = 6$
1^stexcited state : $n_{1} = 2$
Number of spectral lines
$= \frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2} = \frac{(6 - 2) (6 - 2 + 1)}{2} = 2 \times 5 = 10$

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