The energy of an electron in the excited hydrogen atom is $-3.4 \mathrm{eV}$. Then according to Bohr's…
The energy of an electron in the excited hydrogen atom is $-3.4 \mathrm{eV}$. Then according to Bohr's theory, the angular momentum of the electron in that excited state is ( $h=$ Plank's constant $)$
$\frac{2 \pi}{h}$
$\frac{\mathrm{n} h}{2 \pi}$
$\frac{h}{\pi}$
$\frac{3 h}{2 \pi}$
Solution
The energy in ground state $(\mathrm{n}=1)$ is $-13.6 \mathrm{eV}$ The energy in $\mathrm{n}=2$ state is $\frac{-13.6}{4}=-3.4 \mathrm{eV}$ Angular momentum $=\frac{\mathrm{n} h}{2 \pi}=\frac{2 h}{2 \pi}=\frac{h}{\pi}$