Orbital magnetic moment and orbital quantum number An electron with orbital angular momentum quantum number l = 0 has what value of orbital magnetic dipole moment?

Introduction / Context:
Electrons possess both orbital and spin angular momenta. Each angular momentum contributes to a magnetic dipole moment. Recognizing when the orbital contribution vanishes is key in atomic magnetism and in interpreting selection rules and spectral splitting (Zeeman effect).

Given Data / Assumptions:

• Orbital quantum number l specifies the magnitude of orbital angular momentum.
• Magnetic dipole moment due to orbital motion is related to orbital angular momentum.
• Spin magnetic moment is separate and may still be present.

Concept / Approach:

The orbital magnetic dipole moment magnitude is proportional to the orbital angular momentum L. For hydrogenic-like states, |L| = √(l(l + 1)) ħ. When l = 0 (s-orbital), L = 0; therefore, the orbital magnetic dipole moment is zero. This does not preclude a spin magnetic moment, which can still contribute to the atom’s overall magnetic behavior.

Step-by-Step Solution:

Verification / Alternative check:

In atomic spectroscopy, s-electrons show no orbital magnetic splitting from μL, consistent with l = 0. Zeeman splitting patterns corroborate this principle.

Why Other Options Are Wrong:

(b) gives the general l-dependent form but for l = 0 it is zero, not non-zero; (c) equals one Bohr magneton is not generally true; (d) suggests exact cancellation with spin which is not implied; (e) is incorrect since l = 0 is perfectly well-defined.

Common Pitfalls:

Confusing orbital with spin contributions; assuming every electron must have a non-zero orbital moment regardless of l.

Final Answer:

Zero (no orbital magnetic dipole moment)