Question

An electron in a hydrogen atom is in an slevel, and the atom is in a magnetic field BS _ Bkn Explain why the “spin up” state$m_s=+\frac{1}{2}$ has higher energy than the “spin down” state $m_s=\frac{1}{2}$.

Short answer

We get the interaction energy $U=(2.00232){\mu }_{B}B{m}_s$, So, for $m_s=-\frac{1}{2}$, the interaction energy is negative, decreasing the energy of the electron, and for $m_s=+\frac{1}{2}$, the interaction energy is positive, increasing the energy of the electron.

Step-by-step solution

Step 1:

From the below equation, the z-component of the spin angular momentum of the electrons is given by:

$S_z=m_sħ$

…. (1)

From the below equation, the z-component of the spin magnetic moment $u_z$is related to$s_z$ by:

${\mu }_z=-(2.00232)\frac{e}{2m}{S}_z$

Substituting equation (1), we get:

$$\begin{gathered} {\mu }_z=-(2.00232)\frac{e}{2m}\left(m_sħ\right) \\ =-\left(2.00232\right)\frac{eħ}{2m}{m}_s \\ =-\left(2.00232\right){\mu }_B{m}_s \end{gathered}$$

Where${\mu }_B$ is the Bohr magnetron.

Step 2:

From the equation given below, the potential energy U associated with the interaction between a magnetic moment $\overrightarrow{\mu }$and an external magnetic field$\overrightarrow{B}$ is given by:

$U=-\overrightarrow{\mu }\to .\overrightarrow{B}$

But, we have$\overrightarrow{B}=B\stackrel{⏜}{k}$(the magnetic field is in the z-direction only), thus, the interaction is:

$U=-{\mu }_{z}B$

Substituting form equation (2), we get

$U=(2.00232){\mu }_{B}B{m}_s$

So, for$m_s=+\frac{1}{2}$, the interaction energy is:

$U_{1/2}=(1.00116){\mu }_{B}B$

Thus, the energy of this electron increases by this amount

For$m_s=-\frac{1}{2}$, the interaction energy is:

$U_{-1/2}=-(1.00116){\mu }_{B}B$

Thus, the energy of this electron decreases by this amount