Physics | VITEEE 2015 | Discussion Page

The angular momentum of an electron in Bohr 's hydrogen atom whose energy is -3.4 e V , is

$\frac{5h}{2\pi}$

$\frac{h}{2\pi}$

$\frac{h}{\pi}$

$\frac{2h}{3\pi}$

Option C

Energy of electron on the nth orbit of hydrogen atom

$E_{n}=-\frac{13.6}{n^{2}}eV$

$\Rightarrow 3.4=-\frac{13.6}{n^{2}}\Rightarrow n^{2}=4$

or n=2

Angular momentum of electron

$L=\frac{nh}{2\pi}=\frac{2h}{2\pi}=\frac{h}{\pi}$

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