Question

Which energy state of the triply ionized beryllium ($B{e}^{3+}$) has the same electron orbital radius as that of the ground state of hydrogen?

• A. n = 1
• B. n = 2
• C. n = 3
• D. n = 4

Answer 1

The correct option is B n = 2

For an atom of atomic number $Z$, the radius of the $n^{th}$ orbit is given by $r_n=\frac{K{n}^2}{Z}$ where $K$ is a constant
For the ground state of hydrogen, $Z=1,n=1$ and for $B{e}^{3+}$, $Z=4$
By the given condition, $\frac{n_1^2}{Z_1}=\frac{n_2^2}{Z_2}$
$\Rightarrow \frac{1}{1}=\frac{n_2^2}{4}$
$\Rightarrow n_2=2$
Hence, the correct choice is (b).