Question

The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf = 4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ni = 6.

Answer 1

Energy of an electron in a particular state is,\[-\dfrac{Z^2}{n^2}\left(13.6\right) eV\]Helium has \(Z = 2\).

Helium will radiate some energy to get from a higher state to a lower state. This energy will, obviously be, the \(\rm initial~ energy - final~ energy\). The initial energy is in the 6th excited state.\[E_i = -\dfrac{2^2}{6^2}(13.6)eV = \dfrac{-13.6}{9} eV\]Similarly, calculate the energy in the fourth state and subtract that from the initial. That is the energy of the radiation. Also note that \(E = hc/\lambda\).

Answer 2

Alternative: \[\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_{\rm final}^2} - \dfrac{1}{n_{\rm initial}^2}\right)\]\(R\) is Rydberg's Constant.