Find the radius and energy of `He^(+)` ion in the states (a) `n=2, (b) n=3`

Answer

(a) `r=0.529xx(2^(2))/(2)=1.058Å`
`E =-13.6xx(2^(2))/(2^(2))= -13.6 eV`
(b) `r=0.529xx(3^(2))/(2)=2.38Å`
`E = -13.6xx(2^(2))/(3^(2))= -6.04eV`.

Explanation

`r=0.529` ,
(a) `r(n=2)=0.529xx(2^(2))/(2)=1.058Å`
`E(n-2)= -13.6xx(2^(2))/(2^(2))= -13.6 eV`
(b) `r(n=3)= 0.529xx(3^(3))/(2)= 2.38Å`
`E(n=3) = -13.6xx(2^(2))/(3^(2))= -6.04 eV`.