An electron has intrinsic spin angular momentum S=hbar/2. The magnetic moment due to spin is mu_s = g_s,ehbar/2m_e,S/hbar where g_ssimeq2. Compute the value of mu_s in units of Bohr magneton mu_B=ehbar/2m_e.

An electron has intrinsic spin angular momentum $S = \frac{hba r}{2}$ . The magnetic moment due to spin is $m u_{s} = g_{s}, \frac{e hba r}{2 m _{e}}, \frac{S}{hba r}$ where $g_{s} s im e q 2$ . Compute the value of $m u_{s}$ in units of Bohr magneton $m u_{B} = \frac{e hba r}{2 m _{e}}$ .

hard 3 marks Class 12 · Physics · Atoms

Answer

1 μ_B

The correct answer is 1 μ_B — Use $g_{s} s im e q 2$ : $μ_{s} = 2 \times μ_{B} \times \frac{1}{2} = 1, μ_{B}$ .

Express $m u_{s}$ in terms of $m u_{B}$ : $m u_{s} = g_{s}, \frac{e hba r}{2 m _{e}}, \frac{S}{hba r} = g_{s}, m u_{B}, \frac{S}{hba r}$ .
Insert $S = hba r /2$ : $m u_{s} = g_{s}, m u_{B}, \frac{1}{2}$ .
Use $g_{s} s im e q 2$ : $μ_{s} = 2 \times μ_{B} \times \frac{1}{2} = 1, μ_{B}$ .

Common mistake: Using

S = hba r

instead of

hba r /2

Concept tested: Electron spin magnetic moment

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked