Question:easy

According to Bohr's model, the angular momentum of an electron in a stable orbit is an integral multiple of

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Remember that in quantum mechanics, the quantization of angular momentum is expressed as \(L = n \frac{h}{2\pi}\), where \(n\) is a positive integer.
Updated On: Jun 3, 2026
  • $h / 2\pi$
  • $h$
  • $2\pi / h$
  • $h / \pi$
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The Correct Option is A

Solution and Explanation

Step 1: Recall Bohr's idea.
In Bohr's model, electrons can only orbit in certain allowed paths. In these paths the angular momentum takes only special values.

Step 2: Write the quantization rule.
The angular momentum of the electron is a whole-number multiple of $\dfrac{h}{2\pi}$. \[ L = n\frac{h}{2\pi} \] where $n = 1, 2, 3, \dots$ and $h$ is Planck's constant.

Step 3: Identify the basic unit.
The smallest step, when $n = 1$, is \[ L = \frac{h}{2\pi} \] This is the basic packet of angular momentum.

Step 4: Connect to the orbit.
The angular momentum is also $L = mvr$, the mass times speed times radius. Setting this equal to $n\dfrac{h}{2\pi}$ gives the allowed orbits.

Step 5: Rule out the others.
Choices like $h$, $2\pi/h$, or $h/\pi$ do not match the correct unit of $h/2\pi$.

Step 6: State the answer.
The angular momentum is an integer multiple of $h/2\pi$. \[ \boxed{L = n\frac{h}{2\pi}} \]
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