Question

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

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

Answer 1

The correct option is A $1.05\times {10}^{-34}J-s$

Electron after absorbing $10.2eV$ energy goes to its first excited sate $(n=2)$ from ground state $(n=1)$.
$\therefore$ Increase in momentum $=\frac{h}{2\pi }$$=\frac{6.6\times {10}^{-34}}{6.28}=1.05\times {10}^{-34}J-s$