(Solved): 5) The energy level diagram of singly ionized helium is shown below: n=4n=E=3.4eVn=3n=2 ...

5) The energy level diagram of singly ionized helium is shown below: $\begin{array}{lll}\dots & \mathrm{n}=?& \mathrm{E}=0\\ \mathrm{n}=4& \mathrm{E}=?3.4\mathrm{eV}& \\ & \mathrm{n}=3& \mathrm{E}=?6.04\mathrm{eV}\\ & \mathrm{n}=2& \mathrm{E}=?13.6\mathrm{eV}\\ & \mathrm{n}=1& \mathrm{E}=?54.4\mathrm{eV}\end{array}$ While in the $\mathrm{N}$ shell ( $\mathrm{n}=4$ quantum state) the remaining electron makes a spontaneous transition to the $\mathrm{L}$ shell ( $\mathrm{n}=2$ quantum state). $\text{ ? }=6.626\times 10^{?34}\mathrm{Js}\mathrm{c}=3?10^8\mathrm{m}/\mathrm{s}1\mathrm{eV}=1.602?10^{?19}\mathrm{J}$ j. How much energy does the electron lose in this transition? k. What is the wavelength of the photon emitted in this transition? 1. Once in the $\mathrm{n}=2$ quantum state, how much energy must the electron absorb to become free of the atom?