Question

The energy of electron in the excited state of hydrogen atom is $-0.85eV$. If 'h' is Plank's constant, the angular momentum of electron in the excited state is :

• A. $\frac{4h}{2\pi }$
• B. $h$
• C. $\frac{3h}{2\pi }$
• D. $\frac{h}{\pi }$

Answer 1

The correct option is A $\frac{4h}{2\pi }$

Energy of the $nth$ state of H-atom $E_n=\frac{-13.6}{n^2}eV$

So, $-0.85=-\frac{13.6}{n^2}$

Or $n^2=\frac{13.6}{0.85}=16$

$⟹n=4$

Angular momentum $L=\frac{nh}{2\pi }=\frac{4h}{2\pi }$