An alkali metal atom is in the ground state. The orbital angular momentum equals zero and the spin angular momentum is entirely due to the single valence electron. A magnetic field is applied that splits the ground state energy level into two levels, 27 μeV apart. What is the strength of the applied magnetic field? (h = 6.626 × 10^-34 J ∙ s, Bohr magneton = μB = 9.27 × 10^-24 J/T, 1 eV = 1.60 × 10^-19 J)

Question

Quizplus · Accepted Answer

The energy splitting ΔE = 27 μeV = 27 × 10^-6 × 1.60 × 10^-19 J. The splitting is given by ΔE = gμBB, where g ≈ 2 for an electron. Solving for B gives B = ΔE / (gμB) = (27 × 10^-6 × 1.60 × 10^-19) / (2 × 9.27 × 10^-24) ≈ 0.23 T.

An Alkali Metal Atom Is in the Ground State