When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck's constant: $h=6.6 \times 10^{-34}$ J-s)
Answer & explanation
Correct answer: option 3
Change in the angular momentum
$\Delta L=L_2-L_1=\frac{n_2 h}{2 \pi}-\frac{n_1 h}{2 \pi} \Rightarrow \Delta L=\frac{h}{2 \pi}\left(n_2-n_1\right)$
$=\frac{6.6 \times 10^{-34}}{2 \times 3.14}(5-4)=1.05 \times 10^{-34}$ J-s