Number of spectral lines that are emitted by atomic hydrogen excited to the $n t h$ energy level is:

\frac{1}{2} n (n - 1)

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\frac{1}{4} n (n - 1)

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\frac{1}{2} n (n + 1)

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\frac{1}{2} n (n - 2)

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Solution

The correct option is A $\frac{1}{2} n (n - 1)$
Number of spectral lines that are emitted by atomic hydrogen excited to the nth energy level is
$\frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2} = \frac{1}{2} n (n - 1)$
Here $n_{2} = n$ and $n_{1} = 1$

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