An electron in the hydrogen atom excites from 2 nd orbit to 4 th orbit then the change in angular momentum…
An electron in the hydrogen atom excites from 2 nd orbit to 4 th orbit then the change in angular momentum of the electron is
(Planck's constant $h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
$2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
$1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
$0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
$4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
Solution
Change in angular momentum of the electron in $\mathrm{H}$-atom when it is excited from 2nd orbit to 4 th orbit is given as
$\Delta L=L_2-L_1$
Where, $L_2$ is angular momentum of 4 th excited state $(n=5)$ and $L_1$ is angular moment of 2 nd excited ștate $(n=3)$.
$\therefore \quad \Delta L=\frac{n_2 h}{2 \pi}-\frac{n_1 h}{2 \pi}$
$\begin{aligned} & =\frac{h}{2 \pi}\left[n_2-n_1\right]=\frac{6.64 \times 10^{-34}}{2 \times 3.14}[5-4] \\ & =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}\end{aligned}$