Which state of the triply ionized `Be^(+++)` has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states.

Explanation

Radius of nth orbit is given by
`r=(n^2h^2)/(4pi^2mKZe^2) i.e., rprop (n^2)/Z`
for hydrogen, Z=1, n=1 in ground state
`:. (n^2)/Z=(1^2)/1=1`
for Beryleum, Z=4, As orbital radius is
same, `(n^2)/Z=1`
`:. n^2=1xxZ=1xx4=4`
`n=sqrt4=2`
Here, n=2 level of Be has same radius as n=1 level of hydrogen.
Now, energy of electron in nth orbit is
`E=-(2pi^2mK^2Z^2e^4)/(n^2h^2)`
`:. E prop(Z^2)/(n^2)`
`(E_((Be)))/(E_((H)))=([Z^2//n^2]_(Be))/([Z^2//n^2]_H)=(16//4)/(1//1)=4`