Q. When an electron in hydrogen atom is excited, from its $4^{th}$ to 5$^{th}$ stationary orbit, the change in angular momentum of electron is (Planck’s constant: $h = 6.6 \times 10^{- 34}J$-$s$)
A) $4.16 \times 10^{- 34}J$-$s$
B) $3.32 \times 10^{- 34}J$-$s$
C) $1.05 \times 10^{- 34}J$-$s$
D) $2.08 \times 10^{- 34}J$-$s$
Explanation
(c) Change in the angular momentum
$\mathrm{\Delta}L = L_{2} - L_{1} = \frac{n_{2}h}{2\pi} - \frac{n_{1}h}{2\pi} \Rightarrow \mathrm{\Delta}L = \frac{h}{2\pi}(n_{2} - n_{1})$
$= \frac{6.6 \times 10^{- 34}}{2 \times 3.14}(5 - 4) = 1.05 \times 10^{- 34}J - s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.