Q. Energy of an electron in an excited hydrogen atom is $- 3.4\ eV$. Its angular momentum will be $(h = 6.626 \times 10^{- 34}J - s)$
A) $1.11 \times 10^{34}J\ s$
B) $1.51 \times 10^{- 31}J\ s$
C) $2.11 \times 10^{- 34}J\ s$
D) $3.72 \times 10^{- 34}J\ s$
Explanation
(c) The electron is in the second orbit $(n = 2)$
Hence $L = \frac{nh}{2\pi} = \frac{2h}{2\pi} = \frac{6.6 \times 10^{- 34}}{\pi} = 2.11 \times 10^{- 34}J - s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.