Q. What will be the angular momentum of an electron, if energy of this electron in $H$-atom is $- 1.5eV$ (in $J$-$s$)
A) $1.05 \times 10^{- 34}$
B) $2.1 \times 10^{- 34}$
C) $3.15 \times 10^{- 34}$
D) $- 2.1 \times 10^{- 34}$
Explanation
(c) Energy of electron in $H$ atom $E_{n} = \frac{- 13.6}{n^{2}}eV$
$\Rightarrow - 1.5 = \frac{- 13.6}{n^{2}} \Rightarrow n^{2} = \frac{13.6}{1.5} = 3$
Now angular momentum
$L = n\frac{h}{2\pi} = \frac{3 \times 6.6 \times 10^{- 34}}{2 \times 3.14} = 3.15 \times 10^{- 34}J \times s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.