Question

The energy of an excited H-atom is -3.4eV. The angular momentum of $e^{-}$ in the given orbit is:

• A. $h/\pi$
• B. $h/2\pi$
• C. $2h/\pi$
• D. $h/3\pi$

Answer 1

The correct option is A $h/\pi$

As we know,
$E=-13.6\frac{1}{n^2}$;

$3.4=\frac{13.6}{n^2}$; $n=2$

So, angular momentum of e in the given orbit is $mvr=\frac{nh}{2\pi }=\frac{h}{\pi }$ as n is 2 here.