Question

The energy of an electron in a particular orbit of single electron species of beryllium is the same as the energy of an electron in the ground state of a hydrogen atom. Identify the orbit of beryllium.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Energy of an electron in $B{e}^{+3}$ ion
$=\frac{-13.6}{n^2}\times 4^{2}ev$
Energy of electron in ground state hydrogen atom $=-13.6ev$
$\therefore \frac{-13.6}{n^2}\times 4^2=-13.6$
$\Rightarrow n^2=4^2$

$\therefore \frac{-13.6}{4^2}\times 4^2=-13.6$

we get -13.6 on both sides
$\Rightarrow n=4$