Atoms more complicated than hydrogen have more than one proton in their nucleus. Let Z stands for the number of protons in a nucleus. Also imagine that an atom loses all but one of its electrons so that it changes into a positively charged ion with just one electron. Bohr.s formula for the energy levels of the hydrogen atom can easily be changed to apply to such ions. It becomes `E_m = (-Z^2 e^4 m_e)/(8 epsi_0^2 h^2 n^2)`, where m= mass of electron , e= change of electron, n = orbit number.
The ionization potential of Het in ground state is

Options

A. `2427 KJ "mol"^(-1)`
B. `5249 KJ "mol"^(-1)`
C. `7116 KJ "mol"^(-1)`
D. `9811 KJ "mol"^(-1)`

Answer

A

Explanation

`E_n= (kz^2)/(n^2)` , with the help of this, `2^(nd)` ionisation energy of .He. an be calcualted.
`E_(oo) -E_1 =0 - ((-Kz^2)/(I^2)) = 2.179 xx 10^(-18) J xx 4`
`= 8.716 xx 10^(-18) J `
in moles `=8.716 xx 10^(-18) 6.023xx10^(23)`
`=5249 "KJ mol"^(-1)`
first ionisation energy is always lower than second.