Question:
In wave mechanics, the angular momentum of an electron is given by
In wave mechanics, the angular momentum of an electron is given by
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Bohr model gives \(L = n\hbar\), but wave mechanics gives \(L = \sqrt{l(l+1)}\hbar\). These agree only in the classical limit of large \(l\).
Updated On: Apr 20, 2026
- \(\sqrt{l\,\dfrac{h}{2\pi}}\)
- \(\sqrt{l(l+1)}\,\dfrac{h}{2\pi}\)
- \(\sqrt{\dfrac{h2\pi}{l(l+1)}}\)
- \(\sqrt{\dfrac{h2\pi}{l}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Orbital angular momentum is quantised in quantum mechanics. It depends on azimuthal quantum number \(l\).
Step 2: Detailed Explanation:
Quantum mechanical expression: \(L = \sqrt{l(l+1)}\,\hbar\), where \(\hbar = \frac{h}{2\pi}\), \(l = 0,1,2,..........,(n-1)\). This replaces Bohr’s expression: \(L = n\hbar\).
Step 3: Final Answer:
\[ \boxed{L = \sqrt{l(l+1)}\,\frac{h}{2\pi}} \]
Orbital angular momentum is quantised in quantum mechanics. It depends on azimuthal quantum number \(l\).
Step 2: Detailed Explanation:
Quantum mechanical expression: \(L = \sqrt{l(l+1)}\,\hbar\), where \(\hbar = \frac{h}{2\pi}\), \(l = 0,1,2,..........,(n-1)\). This replaces Bohr’s expression: \(L = n\hbar\).
Step 3: Final Answer:
\[ \boxed{L = \sqrt{l(l+1)}\,\frac{h}{2\pi}} \]
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