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.

An electron is present in a hydrogen atom in the ground state and anot

1.	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.
Answer» Solution : 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