What is the energy difference between parallel and antiparallel alignment of the zcomponent of an electron’s spin magnetic dipole moment with an external magnetic field of magnitude0.25 T, directed parallel to the zaxis?

$\overrightarrow{B}=\left(0.25T\right)\hat{Z}$

Potential energy of the spin magnetic dipole moment in an external magnetic field is given by the dot product of spin magnetic dipole moment and external magnetic field. The spin magnetic dipole moment is quantized, and it can take only two discrete values corresponding to parallel or antiparallel orientation of spin.

Formula:

$U=-\overrightarrow{{\mu }_s}·\overrightarrow{B_{ext}}$

For parallel spin state,

$\begin{array}{rcl}{U}_p& =& -\overrightarrow{{\mu }_s}·\overrightarrow{B_{ext}}\\ & =& -\overrightarrow{{\mu }_s}·B\hat{Z}\\ & =& -{{\mu }_s}_z·B\\ & =& \frac{-eh}{4\pi m_e}·\left(0.25\mathrm{T}\right)\end{array}$

For antiparallel spin state,

$\begin{array}{rcl}{U}_a& =& -\overrightarrow{{\mu }_s}·\overrightarrow{B_{ext}}\\ & =& -\overrightarrow{{\mu }_s}·B\hat{Z}\\ & =& -{{\mu }_s}_z·B\\ & =& \frac{+eh}{4\pi m_e}·\left(0.25\right)\end{array}$

The energy difference between the parallel and antiparallel state of electron is

$\Delta U=U_a-U_p$

$\begin{array}{rcl}\Delta U& =& \frac{+eh}{4\pi m_e}·\left(0.25\right)+\frac{eh}{4\pi m_e}·\left(0.25\right)\\ & =& 0.50·\frac{eh}{4\pi m_e}\\ & =& 0.50\times 9.27\times {10}^{-24}\\ & =& 4.6\times {10}^{-24}\mathrm{J}\end{array}$

Thus, the energy difference between parallel and antiparallel state of electron is$\Delta U=4.6\times {10}^{-24}\mathrm{J}$