Compute the first three energy levels of doubly ionized lithium. What is the ionisation potential ?

Explanation

Energy `E_n = -(Z^2 mc^2 alpha^2)/(2n^2)`
For hydrogen `Z = 1 and E_n = (-13.6 eV)/(n^2)`
For lithium , `Z = 3 :. E_n = (-122.4)/(n^2) eV`
`:. E_1 = -122.4 eV`
Ionizaiton potential = 122.4 volts
`E_2 = (-122.4)/4 = -30.8 eV, E_3 = (-122.4)/9 = -13.6 eV, E_4 = (-122.4)/(16) = -7.65 eV`.