Find total energy, `PE & KE` of electron in :
(A) `2^(nd)` orbit of `He^(+)` ion.
(B) `1^(st)` excited state of `Be^(+3)` ion.

Explanation

`E_(n) =- 13.6 xx(Z^(2))/(n^(2)) (eV)/("atom"),E_(n) =- KE = (PE)/(2)`
`(A) E_(2)=-13.6xx(2^(2))/(2^(2))=-13.6" "(B) E_(3)=-13.6xx(4^(2))/(2^(2))=-54.4eV`
`KE = 13.6eV, PE = - 27.2 eV" "KE = 54.4 eV, PE=-108.8eV`