Question

The radius of which of the following orbits is the same as that of the first Bohr's orbit of hydrogen atom?

• A. $H{e}^{+}(n=2)$
• B. $L{i}^{2+}(n=2)$
• C. $L{i}^{2+}(n=3)$
• D. $B{e}^{3+}(n=2)$

Answer 1

The correct option is D $B{e}^{3+}(n=2)$

Bohr's radius for $n^{th}$ orbit, $\left(r_n\right)=0.529\left[\frac{n^2}{Z}\right]\stackrel{\circ }{A}$
where $n=\text{Energy level},Z=\text{Atomic number}$

For $1^{st}$ Bohr orbit of hydrogen atom, $n=1$ and $Z=1$
Radius of $H$ atom $r_H=0.529\stackrel{\circ }{A}$

For $H{e}^{+}$ ion: $n=2,Z=2$
$r_{H{e}^{+}}=0.529\left[\frac{2^2}{2}\right]=2r_H$

For $L{i}^{2+}$ ion: $n=2,Z=3$
$r_{L{i}^{2+}}=0.529\left[\frac{2^2}{3}\right]=\frac{4}{3}{r}_H$

For $L{i}^{2+}$ ion: $n=3,Z=3$
$r_{L{i}^{2+}}=0.529\left[\frac{3^2}{3}\right]=3r_H$

For $B{e}^{3+}$ ion: $n=2,Z=4$
$r_{B{e}^{3+}}=0.529\left[\frac{2^2}{4}\right]=r_H$

$B{e}^{3+}$ (n = 2) has the same radius as that of the $1^{st}$ Bohr’s orbit of hydrogen atom.