Question

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

Answer 1

1. The single electron species is hydrogen that exists in neutral form.
2. For hydrogen ($\mathrm{H}$),$n=1$ and $Z=1$. The radius of the hydrogen atom $r=\frac{k{n}^2}{Z}=k$.
3. The beryllium is an element in the periodic table with an atomic number 4 . For beryllium to be single electron species, three electrons are released to form $B{e}^{3+}$.
4. For berylium, $n=2$ and $Z=4$.
   The radius of $B{e}^{3+}$,
   $=\frac{k{n}^2}{Z}=\frac{k\times 2^2}{4}=k$
5. Thus the distance of the electron in hydrogen ($\mathrm{H}$) and $B{e}^{3+}$ from the nucleus is the same.
6. The mathematical equation to express the energy of an electron is,
   $E_n=-\frac{k{Z}^2}{n^2}$
   The energy of the electron in hydrogen.
   $=-\frac{k{Z}^2}{n^2}=\frac{-k\times 1^2}{1^2}=-k$
   The energy of the electron in ${\mathrm{Be}}^{+3}$
   $=-\frac{k{Z}^2}{n^2}=\frac{-k\times 4^2}{2^2}=-4k$
7. Hence, the difference in energy$=(4k-k)=3k=3\times 13.6\mathrm{eV}=40.8\mathrm{eV}$
8. Finally, the difference in their energies is 40.8 eV.