Question

A hydrogen atom in is ground state absorbs $10.2 \,eV$ of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : $\left(\right.$ Given, Plank's constant $\left.=6.6 \times 10^{-34} \,Js \right)$

• A. $2.10 \times 10^{-34} \,Js$
• B. $1.05 \times 10^{-34} \,Js$
• C. $3.15 \times 10^{-34} \,Js$
• D. $4.2 \times 10^{-34} \, Js$

Answer 1

Answer: (B)

$13.6\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)=10.2$
$n=2$
$L_{i}=\frac{h}{2 \pi} \times 1$
$L_{F}=\frac{2 h}{2 \pi}$
$\Delta L=L_{F}-L_{i}=\frac{h}{2 \pi}=\frac{6.6 \times 10^{-34}}{22}$
$=1.05 \times 10^{-34} J-s$