In the lowest energy level of hydrogen atom, electron has an angular momentum equal to:

\frac{π}{h}

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\frac{h}{π}

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\frac{h}{2 π}

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\frac{2 π}{h}

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Solution

The correct option is C $\frac{h}{2 π}$
The angular momentum $L$ $= m_{e} v r$ is on integer multiple of $\frac{h}{2 π}$
$m v r = \frac{n h}{2 π}$

For, $n = 1$
$m v r = \frac{h}{2 π}$

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