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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.

User Mgrenier
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Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )|
E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J

After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:

E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
User Gregyski
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