An electron is present in a hydrogen atom in the ground state and another electron is present in a smgle electron species of beryllium. In both the species the distance between the nucleus and electren 18 same. Calculate the difference in their energies.

Explanation

radius of hydrogen atom in `1^(st)` orbit
` = (kn^2)/Z=kn^2=k.1^2=k `
Let radius of `n^(th)`orbit of `Be^(+3)=k`
`therefore k=(kn^2)/4, therefore n^2=4, therefore n=2`
ln order to maintain the same distance, the electron has to be present in the `2^(nd)` orbit of `Be^(+3)`. Energy of the electron in the 1st orbit of hydrogen
`= (-kz^2)/(n^2)=-k`
Energy of the electron in the `2^(nd)` orbit of `Be^(+3)`
`=-(kz^2)/(n^2)`
`= -(kxx4xx4)/(2xx2)=-4k`
`therefore` Difference in energy `=(4k-k)=3k`
`=(3xx13.6)eV`
=40.8eV