Question

Question: The most prominent feature of the hydrogen spectrum in the visible region is the red Balmer line, coming from the transition n = 3to n = 2. First of all, determine the wavelength and frequency of this line according to the Bohr Theory. Fine structure splits this line into several closely spaced lines; the question is: How many, and what is their spacing? Hint: First determine how many sublevels the n = 2level splits into, and find ${\mathbf{E}}_{\mathbf{fs}}^{\mathbf{1}}$for each of these, in eV. Then do the same for n = 3. Draw an energy level diagram showing all possible transitions from n = 3to n = 2. The energy released (in the form of a photon) is role="math" localid="1658311193797" $\mathbf{(}{\mathbf{E}}_{\mathbf{3}}\mathbf{-}{\mathbf{E}}_{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{∆}\mathbf{E}$, the first part being common to all of them, and the $\mathbf{∆}\mathbf{E}$(due to fine structure) varying from one transition to the next. Find $\mathbf{∆}\mathbf{E}$(in eV) for each transition. Finally, convert to photon frequency, and determine the spacing between adjacent spectral lines (in Hz- -not the frequency interval between each line and the unperturbed line (which is, of course, unobservable), but the frequency interval between each line and the next one. Your final answer should take the form: "The red Balmer line splits into (???)lines. In order of increasing frequency, they come from the transitionsto (1) j =(???),toj =(???) ,(2) j =(???) to j =(???)……. The frequency spacing between line (1)and line (2)is (???) Hz, the spacing between line (2)and (3) line (???) Hzis……..”

Short answer

The frequency spacing between lines are

$$\begin{gathered} {\nu }_6-{\nu }_5=0.99\times {10}^{9}eV \\ {v}_5-{\nu }_4=3.36\times {10}^{9}eV \\ {\nu }_4-{\nu }_3=6.5\times {10}^{9}eV \\ {\nu }_3-{\nu }_2=1.09\times {10}^{9}eV \\ {\nu }_2-{\nu }_1=3.24\times {10}^{9}eV \end{gathered}$$

Step-by-step solution

Formula Used

To findand, calculate

$\Delta E=h\nu$

$\to \Delta E=E_3^0-E_2^0$

Find the wavelength

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

According to Bohr's theory:

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

where,$E_1=-13.6eV$

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

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

The wavelength is

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

Calculate fine structure

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$=2\Rightarrow l=0,1\Rightarrow j=\frac{1}{2},\frac{3}{2}$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,$$\begin{gathered} E_2=\frac{E_1}{n_2}=\frac{13.6}{4}eV \\ \Rightarrow E_2^{\text{'}}=\frac{{\left(13.6\right)}^2}{{\left(4\right)}^2\left(2\right)(0.511\times {10}^6)}\left(3-8\right)=5.66\times {10}^{-5}eV \\ j=\frac{3}{2}, \\ {E}_2^{,}=\frac{{\left(E_2\right)}^2}{2m{c}^2}\left(3-\frac{8}{{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{2}}}\right) \\ {E}_2^{,}=\frac{{\left(13.6\right)}^2}{{\left(4\right)}^2\left(2\right)\left(0.511\times {10}^6\right)}\left(3-\frac{8}{2}\right) \\ =1.13\times {10}^{-5}eV \end{gathered}$$

- For $n=3\Rightarrow l=0,1,2\Rightarrow j=\frac{1}{2},\frac{3}{2},\frac{5}{2}$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,$E_3=\frac{E_1}{n^2}=\frac{13.6}{9}eV$

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

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

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

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

Then, there are six transitions of energies:

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

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

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

Calculate energy in every transition

Calculatein every transition, then:

1.

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

2.

$$\begin{gathered} \frac{1}{2}\to \frac{3}{2}\Rightarrow ∆E=E_3^{\text{'}}-E_2^{\text{'}} \\ ∆E_4=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511\times {10}^6\right)}\left[\frac{1}{9^2}\left(3-12\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =-8.8\times {10}^{-6}eV \end{gathered}$$

3.

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

4.

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

5.

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

6.

$$\begin{gathered} \frac{5}{2}\to \frac{3}{2}\Rightarrow ∆E=E_3^{\text{'}}-E_2^{\text{'}} \\ ∆E_5=\frac{{\left(13.6\right)}^2}{\left(2\right)\left(0.511\times {10}^6\right)}\left[\frac{1}{9^2}\left(3-4\right)-\frac{1}{4^2}\left(3-4\right)\right] \\ =9.1\times {10}^{-6}eV \end{gathered}$$

Calculate frequency spacing

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

And the other 5-transitions have higher frequencies.

Then, the frequency spacing can be calculated as:

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

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

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

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

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