Question

A hydrogen atom in its ground state absorbs $10.2eV$ of energy. By what amount is the orbital angular momentum increased?

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

Answer 1

Answer: (A)

Energy of electron at ground state $\left(n = 1\right)$ is $-13.6eV$ after absorbing $10.2eV$ energy it total energy becomes $-3.4eV$ . This is energy of first excited state $\left(n = 2\right)$
So, Increase in momentum $=\frac{2 h}{2 \pi }-\frac{h}{2 \pi }=\frac{h}{2 \pi }$
$=\frac{6 . 6 \times 10^{- 34}}{6 . 28}=1.05\times 10^{- 34}J-s$