Question

The series in the He+He+ spectrum that corresponds to the set of transitions where the electron falls from a higher level to the ????f=4nf=4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ????i=8ni=8 .

Answer 1

Explanation:

According to the Rydberg formula,

        $\frac{1}{\lambda} = -R_{H}Z^{2} [\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}]$

where,        $\lambda$ = wavelength

                  $R_{H}$ = Rydberg constant = $1.097 \times 10^{7} m^{-1}$

                  Z = atomic number (for He atomic number is 2)

               $n_{f}$ = 4,      $n_{i}$ = 8

Hence, putting the given values into the above formula as follows.

              $\frac{1}{\lambda} = -R_{H}Z^{2} [\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}]$

                              = $-1.097 \times 10^{7} \times (2)^{2} [\frac{1}{(8)^{2}} - \frac{1}{(4)^{2}}]$    

                             = $0.2056 \times 10^{7} m^{-1}$

                $\lambda = \frac{1}{0.2056 \times 10^{7} m^{-1}}$

                            = $4.86 \times 10^{-7} m$

                            = 486 nm               (as $1 \times 10^{-9}$ = 1 nm)

Thus, we can conclude that the Pickering series wavelength associated with the excited state is 486 nm.