Q. The energy of an electron in an excited hydrogen atom is $- 3.4\ eV$. Its angular momentum is
A) $3.72\ \times \ 10^{- 34}\ Js$
B) $2.11\ \ \times \ 10^{- 34}\ Js$
C) $1.57\ \times \ 10^{- 34}\ Js$
D) $1.11 \times \ 10^{- 34}\ Js$
Explanation
(b) Energy of electron in $n$th energy level in hydrogen atom
=$\frac{- 13.6}{n^{2}}$ $eV$
Here, $\frac{- 13.6}{n^{2}}$ $= - 3.4\ eV$
So, $n$=2
Angular momentum from Bohr’s principle
= n$\frac{h}{2\pi}$=$\frac{2\ \times \ 6.626\ \times 10^{- 34}}{2\ \times 3.14}$
= $2.11$ $\times 10^{- 34}$ $Js$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.