A hydrogen atom initially in an n= 3, l= 1 state makes a transition to the n= 2, l= 0, j= 1/2 state. Find the difference in wavelength between the following two photons: one emitted in a transition that starts in the n= 3, l= 1, j= 3/2 state and one that starts instead in the n= 3, l= 1, j= 1/2 state. Which photon has the longer wavelength?

The energy of a level without any spin orbit coupling is,

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\frac{\mathbf{13}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{e}\mathbf{v}}{{\mathbf{n}}^{\mathbf{2}}}$

When the spin orbit coupling effect is incorporated then,

${\mathbf{E}}_{\mathbf{n}\mathbf{,}\mathbf{j}}\mathbf{=}\mathbf{-}\frac{\mathbf{13}\mathbf{.60}\mathbf{eV}}{{\mathbf{n}}^{\mathbf{2}}}\left[1+\frac{{\alpha }^2}{n^2}\left(\frac{n}{j+\frac{1}{2}}-\frac{3}{4}\right)\right]$ ...(i)

The energy of a photon is equal to the energy difference between transition levels,

$$\begin{gathered} ∆E=\frac{hc}{\lambda } \\ =E_{photon} \end{gathered}$$

For n = 3,

$$\begin{gathered} E_1=-\frac{13.60eV}{3^2} \\ =\frac{-13.60eV}{9} \end{gathered}$$

For, n = 2,

$$\begin{gathered} {\mathrm{E}}_2=-\frac{13.60\mathrm{eV}}{2^2} \\ =\frac{-13.60\mathrm{eV}}{4} \\ \therefore \mathrm{\Delta }E={\mathrm{E}}_2-{\mathrm{E}}_1 \\ \mathrm{\Delta E}=1.889\mathrm{eV} \end{gathered}$$

The wavelength is therefore solved out,

$$\begin{gathered} \lambda =\frac{\mathrm{hc}}{{\mathrm{E}}_{photon}} \\ =\frac{\left(4.136\times {10}^{-15}\mathrm{eV}\cdot \mathrm{s}\right)\left(2.998\times {10}^8\mathrm{m}/\mathrm{s}\right)}{1.889\mathrm{eV}} \\ =6.56\times {10}^{-7}\mathrm{m} \\ =656\mathrm{nm} \end{gathered}$$

The energy difference between the j = 3/2 and j = ½ state is,

$\Delta E_{photon}=E_{3,3/2}-E_{3,1/2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}...\left(ii\right)$

Now, using equation (i), find the energy of the levels,

$$\begin{gathered} j=\frac{3}{2} \\ {E}_{3.3/2}=-\frac{13.60\mathrm{eV}}{3^2}\left[1+\frac{(0.007297{)}^2}{3^2}\left(\frac{3}{\frac{3}{2}+\frac{1}{2}}-\frac{3}{4}\right)\right] \\ =\frac{-13.60\mathrm{eV}}{9}\left(1+4.437\times {10}^{-6}\right) \end{gathered}$$

$$\begin{gathered} j=\frac{1}{2} \\ {E}_{3.3/2}=-\frac{13.60\mathrm{eV}}{3^2}\left[1+\frac{(0.007297{)}^2}{3^2}\left(\frac{3}{\frac{1}{2}+\frac{1}{2}}-\frac{3}{4}\right)\right] \\ =\frac{-13.60\mathrm{eV}}{9}\left(1+13.312\times {10}^{-6}\right) \end{gathered}$$

Put the values in equation (ii),

$$\begin{gathered} {\mathrm{\Delta E}}_{photon}=\left[\frac{\left(13.60\mathrm{eV}\right)}{9}\right]\left(13.312\times {10}^{-6}-4.437\times {10}^{-6}\right) \\ =1.341\times {10}^{-5}\mathrm{eV} \end{gathered}$$

The difference in wavelengths is given as,

$$\begin{gathered} \mathrm{\Delta }\lambda =-\frac{\lambda }{E_{photon}}\mathrm{\Delta }{E}_{photon} \\ =-\frac{656\mathrm{nm}}{1.889\mathrm{eV}}\left(1.341\times {10}^{-5}\mathrm{eV}\right) \\ =-4.66\times {10}^{-3}\mathrm{nm} \\ =-0.00466\mathrm{nm} \end{gathered}$$

So form the energy values it is evident that the photon having transition from j=1/2 has the longest wavelength.