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5 June, 04:46

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.

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  1. 5 June, 06:00
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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
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