A hydrogen atom in its ground state is subjected to an external magnetic field of 1.0 T. What is the energy difference between the spin-up and spin-down states?

External magnetic field(s)" aremagnetic fields that affect an electron microscope from the outside. Measures to prevent an adverse influence of the external magnetic fields on the instrument performance are needed.

This question wants us to calculate the difference in energy between the spin-up and spin-down states when a hydrogen atom in the ground state is subjected to a 1.0 T magnetic field.

The energy $\left(\text{U}\right)$ due to an applied magnetic field is.$-\mu .\mathrm{B}$If we takeB to be in the z direction$\left(BZ\right)$, then $U=-{\mu }_{z}B$.

From equations, we get the orientation energy as follows

$U=\frac{e}{m_e}{m}_{s}hB$ (1)

Where ${\mu }_z=-\frac{e}{m_e}{m}_{s}h$, and ${\text{m}}_{\text{s}}$is the spin quantum number.

Knowing that of an electron can only have two values; $\frac{1}{2}$for spin-up and $-\frac{1}{2}$for a spindown, then the difference in energy can be calculated as follows using equation (1)

$\begin{array}{rcl}\Delta U& =& \frac{e}{2m_e}hB-\frac{-e}{2m_e}hB\\ & =& \frac{e}{2m_e}hB+\frac{e}{2m_e}hB\\ & =& \frac{e}{m_e}hB.....................\left(2\right)\\ & & \end{array}$

Substituting the known values of the constants and B=1.0 T in equation (2), we get

$\begin{array}{rcl}\Delta U& =& \frac{1.6\times {10}^{-19}C}{9.1\times {10}^{-31}kg}\left(1.055\times {10}^{-34}J·s\right)\left(1.0T\right)\\ & =& 1.855\times {10}^{-23}J\left(\frac{1eV}{1.6\times 10-19J}\right)\\ & =& 1.16\times {10}^{-4}eV\end{array}$

Therefore the energy difference between the spin-up and spin-down states is$1.16\times {10}^{-4}eV$