Question

Which state of $B{e}^{3+}$ has the same orbit radius as that of the ground state of hydrogen atom?

• A. $3$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is B $2$

$r_1$ of $H-$atom $=0.53\stackrel{\circ }{\text{A}}$
$r_n$ of $B{e}^{3+}$ atom $=\frac{n^2}{Z}\times r_1(H-\text{atom})$
According to question
$\frac{n^2}{Z}\times r_1(H-\text{atom})=0.53\stackrel{\circ }{\text{A}}\Rightarrow \frac{n^2}{Z}\times 0.53=0.53\stackrel{\circ }{\text{A}}\Rightarrow n^2=Z\Rightarrow n^2=4\Rightarrow n=2$