Q. A hydrogen atom in its ground state absorbs $10.2\ eV$ of energy. The orbital angular momentum is increased by (Given Planck’s constant $h = 6.6 \times 10^{- 34}J - s$)
A) $1.05 \times 10^{- 34}J$-$s$
B) $3.16 \times 10^{- 34}J$-$s$
C) $2.11 \times 10^{- 34}J$-$s$
D) $4.22 \times 10^{- 34}J$-$s$
Explanation
(a) Electron after absorbing $10.2\ eV$ energy goes to its first excited state $(n = 2)$ from ground state $(n = 1)$
$\therefore$ Increase in momentum $= \frac{h}{2\pi}$
$= \frac{6.6 \times 10^{- 34}}{6.28} = 1.05 \times 10^{- 34}J - s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.