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 `therefore e_n = ( - 112.4 )/( n^2) eV `
` therefore E_1 =- 122.4 ` eV
Ionization potential `= 122.4 ` volts
` E_2 =(-122.4 )/( 4 ) =- 30.8eV ,E_3 = (-122.4 )/(9) =- 13.6 eV , E_4 = (-122.5 )/( 16) =- 7.65 eV `