Question

A hydrogen atom in its ground state absorbs $10.2eV$ of energy. The angular momentum of electron of the hydrogen atom will increase by the value of:

(Given, Planck's constant $=6.6\times {10}^{-34}Js$).

• A. $3.15\times {10}^{-34}Js$
• B. $1.05\times {10}^{-34}Js$
• C. $4.2\times {10}^{-34}Js$
• D. $2.10\times {10}^{-34}Js$

Answer 1

The correct option is B $1.05\times {10}^{-34}Js$

Electron goes to its first excited state $(n=2)$ from ground state $(n=1)$ after absorbing $10.2eV$ energy
Alternatively :
$-13.6+10.2=\frac{-13.6}{n^2}$
$\Rightarrow \frac{13.6}{n^2}=3.4$
$\Rightarrow n=2$
Change in Angular momentum:
$\Rightarrow \Delta L=2\times \frac{h}{2\pi }-1\times \frac{h}{2\pi }$
$=\frac{h}{2\pi }$
$\Rightarrow \Delta L\simeq 1.05\times {10}^{-34}Js$