Question

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

(a) 1.05 × 10⁻³⁴ J s
(b) 2.11 × 10⁻³⁴ J s
(c) 3.16 × 10⁻³⁴ J s
(d) 4.22 × 10⁻³⁴ J s

Answer 1

(a) 1.05 × 10⁻³⁴ J s

Let after absorption of energy, the hydrogen atom goes to the nth excited state.

Therefore, the energy absorbed can be written as
$$\begin{gathered} 10.2=13.6\times \left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\ \Rightarrow \frac{10.2}{13.6}=1-\frac{1}{n^2} \\ \Rightarrow \frac{1}{n^2}=\frac{13.6-10.2}{13.6} \\ \Rightarrow \frac{1}{n^2}=\frac{3.4}{13.6} \\ \Rightarrow n^2=4 \\ \Rightarrow n=2 \end{gathered}$$

The orbital angular momentum of the electron in the nth state is given by
$L_{\mathrm{n}}=\frac{nh}{2\pi }$
Change in the angular momentum, $∆L=\frac{2h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$
∴ $∆L=1.05\times {10}^{-34}\mathrm{Js}$