Question:
The orbital magnetic dipole moment of an electron in hydrogen-like atom is \( \sqrt{2} \mu_B \). What is the state of the electron?
The orbital magnetic dipole moment of an electron in hydrogen-like atom is \( \sqrt{2} \mu_B \). What is the state of the electron?
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Magnetic dipole moment depends on the orbital angular momentum quantum number.
Updated On: Mar 26, 2025
- S
- P
- D
- F
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The Correct Option is C
Solution and Explanation
The orbital magnetic dipole moment is given by:
\[ \mu_L = \sqrt{l(l+1)} \mu_B \] For \( \mu_L = \sqrt{2} \mu_B \):
\[ \sqrt{l(l+1)} = \sqrt{2} \] Solving for \( l \), we get \( l = 2 \), which corresponds to the D-state.
\[ \mu_L = \sqrt{l(l+1)} \mu_B \] For \( \mu_L = \sqrt{2} \mu_B \):
\[ \sqrt{l(l+1)} = \sqrt{2} \] Solving for \( l \), we get \( l = 2 \), which corresponds to the D-state.
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