MHT CET · Physics · Atomic Physics
When an electron is excited from its \(4^{\text {th }}\) orbit \(5^{\text {th }}\) stationary orbit, the change in the angular momentum of electron is approximately (Plank's constant \(=\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) )
- A \(2 \times 10^{-34} \mathrm{Js}\)
- B \(6.63 \times 10^{-34} \mathrm{Js}\)
- C \(1 \times 10^{-34} \mathrm{Js}\)
- D \(3.14 \times 10^{-34} \mathrm{Js}\)
Answer & Solution
Correct Answer
(C) \(1 \times 10^{-34} \mathrm{Js}\)
Step-by-step Solution
Detailed explanation
Change in Angular Momentum:
\(\Delta \mathrm{L}=\mathrm{L}_2-\mathrm{L}_1\)
\(=\frac{\mathrm{n}_2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_1 \mathrm{~h}}{2 \pi}\)
\(=\frac{\mathrm{h}}{2 \pi}(5-4)\)
\(=\frac{6.63 \times 10^{-34} \mathrm{Js}}{2 \pi}\)
\(=1 \times 10^{-34} \mathrm{Js}\)
\(\Delta \mathrm{L}=\mathrm{L}_2-\mathrm{L}_1\)
\(=\frac{\mathrm{n}_2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{n}_1 \mathrm{~h}}{2 \pi}\)
\(=\frac{\mathrm{h}}{2 \pi}(5-4)\)
\(=\frac{6.63 \times 10^{-34} \mathrm{Js}}{2 \pi}\)
\(=1 \times 10^{-34} \mathrm{Js}\)
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