Question

A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by

• A. $3.16 \times 10^{−34}$ Js
• B. $1.05 \times 10^{−34}$ Js
• C. $4.22 \times 10^{−34}$ Js
• D. $2.11 \times 10^{−34}$ Js

Answer 1

Answer: (B)

By absorbing 10.2ev, electron goes to $2^{nd} \, orbit \, as \, E^{n}=\frac{-13.6}{n^{2}} ev,$
$E_{1}=-13.6 e V$
$E_{2}=-3.4 e V$
$E_{2}-E_{1}=10.2 e V$
$L_{2}-L_{1}=\frac{n_{2} h}{2 \pi}-\frac{n _{1}h}{2 \pi}=\frac{2 h}{2 \pi}-\frac{h}{2 \pi}=\frac{6.62\times10^{-34}}{2\times3.14}=1.05\times10^{-34} \, J S$