91影视

The given data is shown below,

The number of possible states is 8 which are possible for $n=2$.

Fine structural splitting is shown on the left side. Because of spin-orbit coupling, this splitting happens even in the absence of a magnetic field. The additional Zeeman splitting that happens in the presence of magnetic fields is depicted on the right side.

It is required to determine the states for $n=2$, $(j=\frac{1}{2})$, $l=1$$(j=\frac{1}{2}\text{鈥塷谤}\frac{3}{2})$. Determine all eight states.

$|1鉄�=|2\text{鈥�}0\text{鈥�}\frac{1}{2}\text{鈥�}\frac{1}{2}鉄�$

$|2鉄�=|2\text{鈥�}0\text{鈥�}\frac{1}{2}\text{鈥�}鈭�\frac{1}{2}鉄�$

$|3鉄�=|2\text{鈥�}1\text{鈥�}\frac{1}{2}\text{鈥�}\frac{1}{2}鉄�$

$|4鉄�=|2\text{鈥�}1\text{鈥�}\frac{1}{2}\text{鈥�}鈭�\frac{1}{2}鉄�$

$|5鉄�=|2\text{鈥�}1\text{鈥�}\frac{3}{2}\text{鈥�}\frac{3}{2}鉄�$

$|6鉄�=|2\text{鈥�}1\text{鈥�}\frac{3}{2}\text{鈥�}\frac{1}{2}鉄�$

$|7鉄�=|2\text{鈥�}1\text{鈥�}\frac{3}{2}\text{鈥�}鈭�\frac{1}{2}鉄�$

$|8鉄�=|2\text{鈥�}1\text{鈥�}\frac{3}{2}\text{鈥�}鈭�\frac{3}{2}鉄�$

$E_n$The sum of the fine-structure part $E_{nj}$ and the Zeeman part $E_z$gives the weak-field Zeeman energy.So, it can be represented as follows,

$E=E_{nj}+E_z$

Here,the value of $E_{nj}=\frac{鈭�13.6}{n^2}(1+\frac{{伪}^2}{n^2}(\frac{n}{j+\frac{1}{2}}鈭�\frac{3}{4}))$

Write the value of $E_n$.

$E_n={渭}_n{g}_j{B}_{ext}{m}_j$

Here, ${渭}_n$ is the Bohr magneton, $g_j$isthe Lande g-factor, $B_{ext}$ is the external magnetic field and$n$ isthe principle quantum number.

Write the equation of the Lande$g$-factor.

$g_j=1+\frac{j(j+1)鈭�l(l+1)+\frac{3}{4}}{2j(j+1)}$

Write the value of Bohr magneton.

${渭}_g=\frac{e\mathrm{鈩�}}{2m}$

For each of the eight states, the Lande g-factors are determined as follows:

For first two states,

$\begin{array}{c}{g}_j=1+\frac{\frac{1}{2}(\frac{1}{2}+1)鈭�(0+1)+\frac{3}{4}}{2\left(\frac{1}{2}\right)(\frac{1}{2}+1)}\\ =1+\frac{\frac{3}{4}+\frac{3}{4}}{\frac{3}{2}}\\ =2\end{array}$

For next two states,

$\begin{array}{c}{g}_j=1+\frac{\frac{1}{2}(\frac{1}{2}+1)鈭�1(1+1)+\frac{3}{4}}{2\left(\frac{1}{2}\right)(\frac{1}{2}+1)}\\ =1+\frac{\frac{3}{4}鈭�2+\frac{3}{4}}{\frac{3}{2}}\\ =\frac{2}{3}\end{array}$

For next four steps,

$\begin{array}{c}{g}_j=1+\frac{\left(\frac{3}{2}\right)(\frac{3}{2}+1)鈭�1(1+1)+\frac{3}{4}}{2\left(\frac{3}{2}\right)(\frac{3}{2}+1)}\\ =1+\frac{\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)鈭�2+\frac{3}{4}}{3\left(\frac{5}{2}\right)}\\ =\frac{4}{3}\end{array}$

The Zeeman part will be calculated to get the value of$E_z$.

Write the general equation for Zeeman energy.

.$E_z=g_j{m}_j{渭}_B{B}_{\text{ext}}$

For $|1鉄�$,

$\begin{array}{c}{E}_z=\left(2\right)\left(\frac{1}{2}\right){渭}_B{B}_{ext}\\ ={渭}_B{B}_{ext}\end{array}$

For$|2鉄�$,

$\begin{array}{c}{E}_z=\left(2\right)\left(\frac{鈭�1}{2}\right){渭}_B{B}_{ext}\\ =鈭�{渭}_B{B}_{ext}\end{array}$

For$|3鉄�$,

$\begin{array}{c}{E}_n=\left(\frac{2}{3}\right)\left(\frac{1}{2}\right){渭}_B{B}_{ext}\\ =\frac{1}{3}{渭}_B{B}_{ext}\end{array}$

For$|4鉄�$,

$\begin{array}{c}{E}_z=\left(\frac{2}{3}\right)\left(\frac{鈭�1}{2}\right){渭}_B{B}_{ext}\\ =鈭�\frac{1}{3}{渭}_B{B}_{ext}\end{array}$

For$|5鉄�$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{3}{2}\right){渭}_B{B}_{ext}\\ =2{渭}_B{B}_{ext}\end{array}$

For$|6鉄�$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{1}{2}\right){渭}_B{B}_{ext}\\ =\frac{2}{3}{渭}_B{B}_{ext}\end{array}$

For $|7鉄�$ ,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{鈭�1}{2}\right){渭}_B{B}_{ext}\\ =鈭�\frac{2}{3}{渭}_B{B}_{ext}\end{array}$

For$|8鉄�$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{鈭�3}{2}\right){渭}_B{B}_{ext}\\ =鈭�2{渭}_B{B}_{ext}\end{array}$

Write the general expression for the fine structure.

$E_{nf}=\frac{鈭�13.6}{n^2}(1+\frac{{伪}^2}{n^2}(\frac{n}{j+\frac{1}{2}}鈭�\frac{3}{4}))$

For the first four states, the value of$E_{nf}$is the same that can be represented as follows,

$\begin{array}{c}{E}_{nf}=\frac{鈭�13.6}{2^2}(1+\frac{{伪}^2}{4}(\frac{2}{\frac{1}{2}+\frac{1}{2}}鈭�\frac{3}{4}))\\ =\frac{鈭�13.6}{4}(1+\frac{{伪}^2}{4}(2鈭�\frac{3}{4}))\\ =\frac{鈭�13.6}{4}(1+\frac{{伪}^2}{4}\left(\frac{5}{4}\right))\\ =\frac{鈭�13.6}{4}(1+\frac{5}{16}{伪}^2)\\ =鈭�3.4(1+\frac{5{伪}^2}{16})\end{array}$

For the next four states, the value of $E_{nf}$ is the samethat can be represented as follows,

$\begin{array}{c}{E}_{nf}=\frac{鈭�13.6}{2^2}(1+\frac{{伪}^2}{4}(\frac{2}{\frac{3}{2}+\frac{1}{2}}鈭�\frac{3}{4}))\\ =鈭�3.4(1+\frac{{伪}^2}{4}(1鈭�\frac{3}{4}))\\ =鈭�3.4(1+\frac{{伪}^2}{4}\left(\frac{1}{4}\right))\\ =鈭�3.4(1+\frac{{伪}^2}{16})\end{array}$

Write the expression for the total Zeeman energy.

$E=E_{nj}+E_z$

So, the total Zeeman energy for first four states can be represented as follows,

$\begin{array}{l}{E}_1=鈭�3.4(1+\frac{5{伪}^2}{16})+{渭}_B{B}_{ext}\\ E_2=鈭�3.4(1+\frac{5{伪}^2}{16})鈭�{渭}_B{B}_{ext}\\ E_3=鈭�3.4(1+\frac{5{伪}^2}{16})+\frac{1}{3}{渭}_B{B}_{ext}\\ E_4=鈭�3.4(1+\frac{5{伪}^2}{16})鈭�\frac{1}{3}{渭}_B{B}_{ext}\end{array}$

So, the total Zeeman energy for next four states can be represented as follows,

$\begin{array}{l}{E}_5=鈭�3.4(1+\frac{1}{16}{伪}^2)+2{渭}_B{B}_{ext}\\ E_6=鈭�3.4(1+\frac{1}{16}{伪}^2)+\frac{2}{3}{渭}_B{B}_{ext}\\ E_7=鈭�3.4(1+\frac{1}{16}{伪}^2)鈭�\frac{2}{3}{渭}_B{B}_{ext}\\ E_8=鈭�3.4(1+\frac{1}{16}{伪}^2)鈭�2{渭}_B{B}_{ext}\end{array}$

Thus, only two energy values, $E_{nj}$are used to indicate the total energy values on the graph, and splitting occurs with an increase of $B_{ext}$.

The representation of the total energies on graph is as follows,