91影视

To findand, calculate

$螖E=h谓$

$鈫�螖E=E_3^0-E_2^0$

There are transition from n = 3 to n=2 ,

According to Bohr's theory:

$$\begin{gathered} 鈭�E=E_3^0-E_2^0 \\ =E_1\left(\frac{1}{n_3^2}-\frac{1}{n_2^2}\right) \\ =E_1\left(\frac{1}{9}-\frac{1}{4}\right) \end{gathered}$$

where,$E_1=-13.6eV$

then,$$\begin{gathered} 鈭�E=-13.6\left(\frac{1}{9}-\frac{1}{4}\right) \\ =1.889eV \end{gathered}$$

$$\begin{gathered} h=\frac{6.62脳10}{1.6脳{10}^{19}} \\ =4.14脳{10}^{-15}eV/s \\ 鈭�E=hv \\ v=\frac{鈭�E}{h} \\ =\frac{1.889}{4.14脳{10}^{-15}} \\ =4.56脳{10}^{14}\mathrm{Hz} \end{gathered}$$

The wavelength is

$$\begin{gathered} 位=\frac{c}{v} \\ =\frac{3脳{10}^8}{4.56脳{10}^{14}} \\ =6.57脳{10}^{-7}m \end{gathered}$$

To calculate the fine structure:

$E_{fs}^{\text{'}}=\frac{{\left(En\right)}^2}{2m{c}^2}\left(3-\frac{4n}{j+{\displaystyle \frac{1}{2}}}\right)$

- For$=2鈬�l=0,1鈬�j=\frac{1}{2},\frac{3}{2}$we have 2 -levels due to the split of n = 2,

$$\begin{gathered} j=\frac{1}{2} \\ {E}_2^{\text{'}}=\frac{{\left(E^2\right)}^2}{2m{c}^2}\left(3-\frac{8}{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right) \end{gathered}$$

Where,$$\begin{gathered} E_2=\frac{E_1}{n_2}=\frac{13.6}{4}eV \\ 鈬�E_2^{\text{'}}=\frac{{\left(13.6\right)}^2}{{\left(4\right)}^2\left(2\right)(0.511脳{10}^6)}\left(3-8\right)=5.66脳{10}^{-5}eV \\ j=\frac{3}{2}, \\ {E}_2^{,}=\frac{{\left(E_2\right)}^2}{2m{c}^2}\left(3-\frac{8}{{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{2}}}\right) \\ {E}_2^{,}=\frac{{\left(13.6\right)}^2}{{\left(4\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-\frac{8}{2}\right) \\ =1.13脳{10}^{-5}eV \end{gathered}$$

- For $n=3鈬�l=0,1,2鈬�j=\frac{1}{2},\frac{3}{2},\frac{5}{2}$we have 3-levels splitting for n =3

$j=\frac{1}{2}$

$E_3^{,}=\frac{{\left(E_3\right)}^2}{2m{c}^2}\left(3-\frac{12}{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right)$

Where,$E_3=\frac{E_1}{n^2}=\frac{13.6}{9}eV$

$$\begin{gathered} 鈬�E_3^{\text{'}}=\frac{E_1}{n^2}=\frac{13.6}{9}eV \\ 鈬�E_3^{\text{'}}=\frac{{\left(13.6\right)}^2}{{\left(9\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-12\right)=-2.01脳{10}^{-5}eV \\ j=\frac{3}{2} \end{gathered}$$

$$\begin{gathered} E_3^{\text{'}}=\frac{{\left(E_3\right)}^2}{2m{c}^2}\left(3-\frac{12}{{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{2}}}\right) \\ {E}_3^{\text{'}}=\frac{{\left(13.6\right)}^2}{{\left(9\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-6\right) \\ =-6.7脳{10}^{-6}eV \end{gathered}$$

For$n=3鈬�l=0,1,2鈬�j=\frac{1}{2},\frac{3}{2},\frac{5}{2}$

localid="1658383160262" $$\begin{gathered} j=\frac{5}{2} \\ {E}_3^{\text{'}}=\frac{{\left(E_3\right)}^2}{2m{c}^2}\left(3-\frac{12}{{\displaystyle \frac{5}{2}}+{\displaystyle \frac{1}{2}}}\right) \\ {E}_3^{\text{'}}=\frac{{\left(13.6\right)}^2}{{\left(9\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-4\right) \\ =2.23脳{10}^{6}eV \end{gathered}$$

Then, there are six transitions of energies:

$鈭�E=\left(E_3^0+E_3^{\text{'}}\right)-\left(E_2^0+E_2^{\text{'}}\right)=\left(E_3^0-E_2^0\right)-\left(E_3^{\text{'}}-E_2^{\text{'}}\right)$

We need to calculate$\left(E_3^{\text{'}}+E_2^{\text{'}}\right)$$\left(E_3^{\text{'}}+E_2^{\text{'}}\right)$,

Take$鈭�E=E_3^{\text{'}}+E_2^{\text{'}}$

Calculatein every transition, then:

1.

$$\begin{gathered} E_3^{\text{'}}=\frac{{\left(E_3\right)}^2}{2m{c}^2}\left(3-\frac{12}{j+{\displaystyle \frac{1}{2}}}\right) \\ =\frac{{\left(E_1\right)}^2}{{\left(9\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-\frac{12}{j+{\displaystyle \frac{1}{2}}}\right) \\ {E}_2^{\text{'}}=\frac{{\left(E_2\right)}^2}{2m{c}^2}\left(3-\frac{8}{j+{\displaystyle \frac{1}{2}}}\right) \\ =\frac{{\left(E_1\right)}^2}{{\left(4\right)}^2\left(2\right)\left(0.511脳{10}^6\right)}\left(3-\frac{8}{j+{\displaystyle \frac{1}{2}}}\right) \\ 鈭�E_4=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-12\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =-3.6脳{10}^{-5}eV \end{gathered}$$

2.

$$\begin{gathered} \frac{1}{2}鈫�\frac{3}{2}鈬�鈭�E=E_3^{\text{'}}-E_2^{\text{'}} \\ 鈭�E_4=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-12\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =-8.8脳{10}^{-6}eV \end{gathered}$$

3.

$$\begin{gathered} \frac{3}{2}鈫�\frac{1}{2}鈬�鈭�E=E_3^{\text{'}}-E_2^{\text{'}} \\ 鈭�E_5=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-12\right)-\frac{1}{4^2}\left(3-8\right)\right] \\ =-4.99脳{10}^{-5}eV \end{gathered}$$

4.

localid="1658384765193" $$\begin{gathered} \frac{3}{2}鈫�\frac{3}{2}鈬�鈭�E=E_3^{\text{'}}-E_2^{\text{'}} \\ 鈭�E_2=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-6\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =-4.6脳{10}^{-6}eV \end{gathered}$$

5.

$$\begin{gathered} \frac{5}{2}鈫�\frac{1}{2}鈬�鈭�E=E_3^{\text{'}}-E_2^{\text{'}} \\ 鈭�E_6=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-4\right)-\frac{1}{4^2}\left(3-8\right)\right] \\ =-5.4脳{10}^{-5}eV \end{gathered}$$

6.

$$\begin{gathered} \frac{5}{2}鈫�\frac{3}{2}鈬�鈭�E=E_3^{\text{'}}-E_2^{\text{'}} \\ 鈭�E_5=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511脳{10}^6\right)}\left[\frac{1}{9^2}\left(3-4\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =9.1脳{10}^{-6}eV \end{gathered}$$

The transition $\left(\frac{1}{2}鈫�\frac{3}{2}\right)$has frequency less than the unperturbed line,

And the other 5-transitions have higher frequencies.

Then, the frequency spacing can be calculated as:

$$\begin{gathered} v_6-v_5=\frac{鈭�E_6}{h}-\frac{鈭�E_6}{h} \\ =\frac{1}{4.14脳{10}^{-15}}\left(5.4脳{10}^{-5}-4.99脳{10}^{-5}\right) \\ =0.99脳{10}^{9}eV \end{gathered}$$

$$\begin{gathered} v_5-v_4=\frac{鈭�E_5}{h}-\frac{鈭�E_4}{h} \\ =\frac{1}{4.14脳{10}^{-15}}\left(4.99脳{10}^{-5}-3.6脳{10}^{-5}\right) \\ =3.36脳{10}^{9}eV \end{gathered}$$

role="math" localid="1658385405705" $$\begin{gathered} v_4-v_3=\frac{鈭�E_4}{h}-\frac{鈭�E_3}{h} \\ =\frac{1}{4.14脳{10}^{-15}}\left(3.6脳{10}^{-5}-9.1脳{10}^{-6}\right) \\ =6.5脳{10}^{9}eV \end{gathered}$$

$$\begin{gathered} v_3-v_2=\frac{鈭�E_3}{h}-\frac{鈭�E_2}{h} \\ =\frac{1}{4.14脳{10}^{-15}}\left(9.1脳{10}^{-6}-4.6脳{10}^{-6}\right) \\ =1.09脳{10}^{9}eV \end{gathered}$$

$$\begin{gathered} v_2-v_1=\frac{鈭�E_2}{h}-\frac{鈭�E_1}{h} \\ =\frac{1}{4.14脳{10}^{-15}}\left(4.6脳{10}^{-6}+8.8脳{10}^{-6}\right) \\ =3.24脳{10}^{9}eV \end{gathered}$$