Question

If the energy of the electron in hydrogen atom in a certain excited state is $-3.4eV$, then what will be its angular momentum?

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

Answer 1

The correct option is B $\frac{h}{\pi }$

Energy of an $e^{-}$ in any excited state is given by:
$E_n=-13.6\frac{Z^2}{n^2}\text{eV/atom}$
For $H$ atom, $E_n=\frac{-13.6}{n^2}\text{eV/atom}$

Since, $E_n=-3.4\text{eV/atom}$ is given, the value of 'n' corresponding to this value from above expression will be $n=2$.
Now,
$\text{Angular momentum}=m\text{v}r=\frac{nh}{2\pi }=\frac{h}{\pi }$