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.

In order to solve this problem, you need the commonly used equation that describes the hydrogen atom. Then you simply need to modify that equation in order to account for the differences in proton number between helium and hydrogen. For hydrogen, an energy level has energy (in eV) given by the following equation: \[E = \frac{-13.6}{n^{2}}\] Any other element has more than 1 proton and 1 electron, so we need to generalize the expression above. I am not going to prove this, but you find that: \[E=\frac{-13.6}{n^{2}}Z^{2}\]Z in this equation is the proton number. For helium, this would be 2. To solve your problem, you need to calculate the energy at each of the levels you are given (n=4 and n=6). Then use the equation which relates energy and wavelength to find the wavelength of the transition: \[E = \frac{hc}{\lambda}\] You may also wish (if the above seems too long) to just use the Rydberg equation, keeping in mind the modification needed ( R is Rydberg's constant). For hydrogen: \[\frac{1}{\lambda}=R(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})\] So for helium: \[\frac{1}{\lambda}=Z^{2}R(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})\] Additionally, you will require use of the following conversion: 1 eV = 1.6x10^-19 J

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