Explanation:
According to the Rydberg formula,
[tex]\frac{1}{\lambda} = -R_{H}Z^{2} [\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}][/tex]
where, [tex]\lambda[/tex] = wavelength
[tex]R_{H}[/tex] = Rydberg constant = [tex]1.097 \times 10^{7} m^{-1}[/tex]
Z = atomic number (for He atomic number is 2)
[tex]n_{f}[/tex] = 4, [tex]n_{i}[/tex] = 8
Hence, putting the given values into the above formula as follows.
[tex]\frac{1}{\lambda} = -R_{H}Z^{2} [\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}][/tex]
= [tex]-1.097 \times 10^{7} \times (2)^{2} [\frac{1}{(8)^{2}} - \frac{1}{(4)^{2}}][/tex]
= [tex]0.2056 \times 10^{7} m^{-1}[/tex]
[tex]\lambda = \frac{1}{0.2056 \times 10^{7} m^{-1}}[/tex]
= [tex]4.86 \times 10^{-7} m[/tex]
= 486 nm (as [tex]1 \times 10^{-9}[/tex] = 1 nm)
Thus, we can conclude that the Pickering series wavelength associated with the excited state is 486 nm.
