The electron is present in the $- 1.51$ eV energy state. Find the angular momentum of the electron.

\frac{h}{π}

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\frac{h}{2 π}

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\frac{3 h}{2 π}

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\frac{2 h}{π}

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none of these

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Solution

The correct option is B $\frac{3 h}{2 π}$
Energy of hydrogen electron in $n^{t h}$ orbit is $- 13.6 / n^{2} = - 1.51 \Rightarrow n = 3$
Now, angular momentum is $n h = 3 h$

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