An electron has intrinsic spin angular momentum S=ħ/2. The spin magnetic moment is μ_s = -g_s e/2m_eS. Given g_s=2.0023, calculate |μ_s| in units of the Bohr magneton μ_B = ehbar/2m_e.

An electron has intrinsic spin angular momentum $S = \frac{ℏ}{2}$ . The spin magnetic moment is $μ_{s} = - g_{s} \frac{e}{2 m _{e}} S$ . Given $g_{s} = 2.0023$ , calculate $∣ μ_{s} ∣$ in units of the Bohr magneton $μ_{B} = \frac{e hba r}{2 m _{e}}$ .

medium 2 marks Class 12 · Physics · Atoms

Answer

1.001 μ_B

The correct answer is 1.001 μ_B — Insert $g_{s} = 2.0023$ : $∣ m u_{s} ∣ = 1.00115, m u_{B} a pp r o x 1.001, m u_{B}$ .

Substitute $S = \frac{ℏ}{2}$ into $μ_{s}$ : $μ_{s} = - g_{s} \frac{e}{2 m _{e}} \cdot \frac{ℏ}{2} = - g_{s} \frac{e hba r}{4 m _{e}} = - \frac{g _{s}}{2}, μ_{B}$ .
Take the magnitude: $∣ μ_{s} ∣ = \frac{g _{s}}{2}, μ_{B}$ .
Insert $g_{s} = 2.0023$ : $∣ m u_{s} ∣ = 1.00115, m u_{B} a pp r o x 1.001, m u_{B}$ .

Common mistake: Forgetting the factor 1/2 that arises from

S = hba r /2

Concept tested: Spin magnetic moment

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked