Question:
According to Bohr's model, the angular momentum of an electron in a stable orbit is an integral multiple of
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: Concept
In Bohr's model of the atom, electrons are assumed to orbit the nucleus in specific, quantized orbits. The angular momentum \(L\) of an electron in these orbits is given by a formula that involves Planck's constant \(\hbar\), which is defined as \(h / 2\pi\).
Step 2: Meaning
The statement means that the angular momentum \(L\) of an electron in any stable orbit according to Bohr's model can be expressed as: \[L = mvr = n \frac{h}{2\pi}\] where \(m\) is the mass of the electron, \(v\) is its velocity, \(r\) is the radius of the orbit, and \(n\) is a positive integer representing the quantum number.
Step 3: Analysis
To prove that the correct answer is A) \(h / 2\pi\), we need to consider the quantization condition for angular momentum in Bohr's model. The angular momentum \(L\) must be an integral multiple of \(\frac{h}{2\pi}\). This can be derived from the relation: \[L = n \frac{h}{2\pi}\] where \(n\) is a positive integer (1, 2, 3, ...). Let's analyze each option: Option A) \(h / 2\pi\): This directly matches the quantization condition for angular momentum in Bohr's model. Option B) \(h\): While Planck's constant \(h\) is involved, it does not match the form required by Bohr's model. Option C) \(2\pi / h\): This is the inverse of \(\frac{h}{2\pi}\), which does not fit the quantization condition. Option D) \(h / \pi\): This is also not in the correct form as it involves \(\pi\) instead of \(2\pi\).
Step 4: Conclusion
The angular momentum of an electron in a stable orbit according to Bohr's model must be an integral multiple of \(\frac{h}{2\pi}\).
Final Answer: (A)
In Bohr's model of the atom, electrons are assumed to orbit the nucleus in specific, quantized orbits. The angular momentum \(L\) of an electron in these orbits is given by a formula that involves Planck's constant \(\hbar\), which is defined as \(h / 2\pi\).
Step 2: Meaning
The statement means that the angular momentum \(L\) of an electron in any stable orbit according to Bohr's model can be expressed as: \[L = mvr = n \frac{h}{2\pi}\] where \(m\) is the mass of the electron, \(v\) is its velocity, \(r\) is the radius of the orbit, and \(n\) is a positive integer representing the quantum number.
Step 3: Analysis
To prove that the correct answer is A) \(h / 2\pi\), we need to consider the quantization condition for angular momentum in Bohr's model. The angular momentum \(L\) must be an integral multiple of \(\frac{h}{2\pi}\). This can be derived from the relation: \[L = n \frac{h}{2\pi}\] where \(n\) is a positive integer (1, 2, 3, ...). Let's analyze each option: Option A) \(h / 2\pi\): This directly matches the quantization condition for angular momentum in Bohr's model. Option B) \(h\): While Planck's constant \(h\) is involved, it does not match the form required by Bohr's model. Option C) \(2\pi / h\): This is the inverse of \(\frac{h}{2\pi}\), which does not fit the quantization condition. Option D) \(h / \pi\): This is also not in the correct form as it involves \(\pi\) instead of \(2\pi\).
Step 4: Conclusion
The angular momentum of an electron in a stable orbit according to Bohr's model must be an integral multiple of \(\frac{h}{2\pi}\).
Final Answer: (A)
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