KCET · Physics · Atomic Physics
A hydrogen atom in ground state absorbs \( 10.2 \mathrm{eV} \) of energy. The orbital angular momentum of
the electron is increased by
- A \( 3.16 \times 10^{-34} \mathrm{Js} \)
- B \( 1.05 \times 10^{-34} \mathrm{Js} \)
- C \( 4.22 \times 10^{-34} \mathrm{Js} \)
- D \( 2.11 \times 10^{-34} \mathrm{~J}_{S} \)
Answer & Solution
Correct Answer
(B) \( 1.05 \times 10^{-34} \mathrm{Js} \)
Step-by-step Solution
Detailed explanation
(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
\[
\begin{array}{l}
E_{1}=-13.6 \mathrm{eV} \\
E_{2}=-3.4 \mathrm{eV} \\
E_{2}-E_{1}=10.2 \mathrm{eV} \\
L_{2}-L_{1}=\frac{n_{2} h}{2 \pi}-\frac{n_{1} h}{2 \pi}=\frac{2 h}{2 \Pi}-\frac{h}{2 \pi}=\frac{6.62 \times 10^{-34}}{2 \times 3.14} \\
=1.05 \times 10^{-34} \mathrm{Js}
\end{array}
\]
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