Ionisation energy of hydrogen atom is 13.6eV.Calculate the ionisation energy for `Be^(3+)` in the first excited sate.

Answer

`122.4eV`

Explanation

Inoization energy of hydrogen `=-E_(1)`(Energy of first Bohr orbit )
` I.E.=-E_(1)=-[-13.6(Z^(2))/(n^(2)]]=+13.6xx(1^(2))/1^(2)=13.6eV`
Ionization energy of `Be^(3+)`
` Z=4
n=2(for 1^(st) "execited state")`
`I.E=[13.6(4^(2))/(2^(2)]]=54.4ev `