Angular Momenta and Magnetic Dipole Moments — Question 2

Question 2

The lowest electron spin energy has spin opposite to the field

For an electron use $\boldsymbol\mu_s=-g_s\mu_B\boldsymbol S/\hbar$, $g_s=2.00$, and $U=-\mu_zB$. A field $B=.800\,\mathrm T$ points along positive z. Use $\mu_B=5.788\times10^{-5}\,\mathrm{eV/T}$.

1. Find both spin energies and identify the lower state.

2. Find the energy required to flip a lower-state electron.

Question 2 - Solution

The lowest electron spin energy has spin opposite to the field

(a) Method and calculation

$$U(m_s)=g_s\mu_BBm_s,\quad m_s=\pm\tfrac12.$$$$\boxed{U_\pm=\pm4.6304\times10^{-5}\,\mathrm{eV}};\quad m_s=-1/2\text{ is lower}.$$

(b) Interpretation and further calculation

$$\Delta U=U_+-U_-=g_s\mu_BB.$$$$\boxed{\Delta U=9.2608\times10^{-5}\,\mathrm{eV}}.$$

Physical check

The lower-energy magnetic moment points along the field, but electron spin points opposite its moment. This sign distinction matters when labeling Zeeman levels.