91影视

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

$S_z=m_s魔$

鈥�. (1)

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

${渭}_z=-(2.00232)\frac{e}{2m}{S}_z$

Substituting equation (1), we get:

$$\begin{gathered} {渭}_z=-(2.00232)\frac{e}{2m}\left(m_s魔\right) \\ =-\left(2.00232\right)\frac{e魔}{2m}{m}_s \\ =-\left(2.00232\right){渭}_B{m}_s \end{gathered}$$

Where${渭}_B$ is the Bohr magnetron.

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

$U=-\stackrel{鈫�}{渭}鈫�.\stackrel{鈫�}{B}$

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

$U=-{渭}_{z}B$

Substituting form equation (2), we get

$U=(2.00232){渭}_{B}B{m}_s$

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

$U_{1/2}=(1.00116){渭}_{B}B$

Thus, the energy of this electron increases by this amount

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

$U_{-1/2}=-(1.00116){渭}_{B}B$

Thus, the energy of this electron decreases by this amount