Question:

The difference in magnitudes of angular momentum of the electrons revolving in \(5^{\text{th}}\) Bohr’s orbit and \(3^{\text{rd}}\) Bohr’s orbit of hydrogen atom is

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Bohr quantization: - $L = n\frac{h}{2\pi}$ - Difference depends only on change in $n$
Updated On: Apr 30, 2026
  • $\frac{2h}{\pi}$
  • $\frac{h}{\pi}$
  • $\frac{h}{2\pi}$
  • $\frac{3h}{2\pi}$
  • $\frac{5h}{2\pi}$
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The Correct Option is B

Solution and Explanation

Concept: In Bohr’s model: \[ L_n = n\frac{h}{2\pi} \]

Step 1:
Find angular momentum for $n=5$.
\[ L_5 = 5\frac{h}{2\pi} \]

Step 2:
Find angular momentum for $n=3$.
\[ L_3 = 3\frac{h}{2\pi} \]

Step 3:
Find difference.
\[ \Delta L = L_5 - L_3 = (5 - 3)\frac{h}{2\pi} = 2\frac{h}{2\pi} = \frac{h}{\pi} \]
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