Question

A hydrogen atom in the ground state emits a photon of energy 12.1 eV. Its orbital angular momentum changes by $\Delta$L. Then $\Delta$L is equal to

• 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

The correct option is B $2.11\times {10}^{-34}Js$

$\Delta E=12.1=E_1-E_n$
$E_n=13.6eV-12.1$eV
$E_n=1.5$eV
$\Rightarrow n=3$
$L_1=\frac{2}{2\pi }$
$L_3=\frac{3h}{2\pi }$
$\Delta L=L_3-L_1=\frac{3h}{2\pi }-\frac{h}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }=\frac{6.6}{3.14}\times {10}^{-34}{s}^{-1}$
$\Delta L=2.11\times 0^{-34}Js$