A hydrogen atom in ground state absorbs 10.2 eV of energy. The or…

MCQ

A hydrogen atom in ground state absorbs $10.2 eV$ of energy. The orbital angular momentum of the electron is increased by:

✓
$1.05 \times 10^{-34} \mathrm{Js}$
B
$2.11 \times 10^{-34} \mathrm{Js}$
C
$3.16 \times 10^{-34} \mathrm{Js}$
D
$4.22 \times 10^{-34} \mathrm{Js}$

✓

Answer

Correct option: A.

$1.05 \times 10^{-34} \mathrm{Js}$

Let after absorption of energy, the hydrogen atom goes to the $n^{th}$ excited state.
Therefore, the energy absorbed can be written as,
$10.2=13.6\times\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$
$\Rightarrow\frac{10.2}{13.6}=1-\frac{1}{\text{n}^2}$
$\Rightarrow\frac{1}{\text{n}^2}=\frac{13.6-10.2}{13.6}$
$\Rightarrow\frac{1}{\text{n}^2}=\frac{3.4}{13.6}$
$\Rightarrow\text{n}^2=4$
$\Rightarrow\text{n}=2$
The orbital angular momentum of the electron in the $n^{th}$ state is given by,
$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Change in the angular momentum, $\Delta\text{L}=\frac{2\text{h}}{2\pi}-\frac{\text{h}}{2\pi}=\frac{\text{h}}{2\pi}$
$\therefore\Delta\text{T}=1.05\times10^{-34}\text{Js}$

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