Q138 / 180 · KCET 2019 · Physics
A hydrogen atom in ground state absorbs \( 10.2 \mathrm{eV} \) of energy. The orbital angular momentum of
the electron is increased by
the electron is increased by
- A\( 3.16 \times 10^{-34} \mathrm{Js} \)
- Correct answerB\( 1.05 \times 10^{-34} \mathrm{Js} \)
- C\( 4.22 \times 10^{-34} \mathrm{Js} \)
- D\( 2.11 \times 10^{-34} \mathrm{~J}_{S} \)
Solution
(B)
By absorbing \( 10.2 \mathrm{ev} \), electron goes to \( 2 \) nd orbit as
\[ \begin{array}{l} E_{1}=-13.6 \mathrm{eV} \\ E_{2}=-3.4 \mathrm{eV} \\ E_{2}-E_{1}=10.2 \mathrm{eV} \\ L_{2}-L_{1}=\frac{n_{2} h}{2 \pi}-\frac{n_{1} h}{2 \pi}=\frac{2 h}{2 \Pi}-\frac{h}{2 \pi}=\frac{6.62 \times 10^{-34}}{2 \times 3.14} \\ =1.05 \times 10^{-34} \mathrm{Js} \end{array} \]
By absorbing \( 10.2 \mathrm{ev} \), electron goes to \( 2 \) nd orbit as
\[ \begin{array}{l} E_{1}=-13.6 \mathrm{eV} \\ E_{2}=-3.4 \mathrm{eV} \\ E_{2}-E_{1}=10.2 \mathrm{eV} \\ L_{2}-L_{1}=\frac{n_{2} h}{2 \pi}-\frac{n_{1} h}{2 \pi}=\frac{2 h}{2 \Pi}-\frac{h}{2 \pi}=\frac{6.62 \times 10^{-34}}{2 \times 3.14} \\ =1.05 \times 10^{-34} \mathrm{Js} \end{array} \]
Hence, the correct answer is option: 'B'.