Question

When a hydrogen atom emits a photon of energy $12.1eV$, its orbital angular momentum changes by

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

Answer 1

Answer: (B)

$2.11\times {10}^{-34}$Js

When a hydrogen atom emits a photon of energy 12.1 eV, i.e. the energy difference between two levels hence,
$E_{n_2}-E_{n_1}=12.1$
$E_{n_2}=12.1+E_{n_1}$
$E_{n_2}=12.1+(-13.6)$
$E_{n_2}=-1.5eV$

This energy corresponds to the third orbit hence, change in orbital momentum is:
$\Delta L=\frac{h}{2\pi }(n_2-n_1)$

$\Delta L=\frac{h}{2\pi }(3-1)=\frac{h}{\pi }$

$\Delta L=\frac{6.6\times {10}^{-34}}{3.14}$

$\Delta L=2.11\times {10}^{-34}Js$