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What will be the angular momentum of a electron, if energy of this electron in H-atom is \mathrm{1.5\, eV\left ( in\: J-sec \right )}

Option: 1

1.05 \times 10^{-34}


Option: 2

2.1 \times 10^{-34}


Option: 3

3.15 \times 10^{-34}


Option: 4

-2.1 \times 10^{-34}


Answers (1)

best_answer

Energy of electron in \mathrm{H} atom \mathrm{E_n=\frac{-13.6}{n^2} \mathrm{eV}}

\mathrm{\Rightarrow-1.5=\frac{-13.6}{n^2} \Rightarrow n^2=\frac{13.6}{1.5}=3}

Now angular momentum
\mathrm{p=n \frac{h}{2 \pi}=\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}=3.15 \times 10^{-34} J \times \sec}

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