The second ionization energy of He is almost exactly four times the ionization energy of H, and the third ionization energy of Li is almost exactly nine times the ionization energy of H:

Question

IE (MJ mol-1)
H(g) --> H +(g) + e- (1.3120)
He+(g) --> He2+(g) + e- (5.2504)
Li2+(g) --> Li3+(g) + e- (11.8149)

Explain this trend on the basis of the Bohr equation for energy levels of single-electron systems.

Answer 1

The Bohr model is a function of RZ^2(1/n^2 - 1/n^2) where the first n is n1 and the second is n2
Z is the atomic number.
He is 2 and 2^2 = 4
Li is 3 and 3^2 = 9