Question

The ratio of radii of first Bohr orbits of $H{e}^{+},L{i}^{2+}$ and $B{e}^{3+}$ is:

• A. $6:4:3$
• B. $6:3:2$
• C. $2:6:3$
• D. $3:2:6$

Answer 1

The correct option is A $6:4:3$

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

For $H{e}^{+}$ ion: $n=1,Z=2$
$r_{H{e}^{+}}=0.529\left[\frac{1^2}{2}\right]\stackrel{\circ }{A}$
For $L{i}^{2+}$ ion: $n=1,Z=3$
$r_{L{i}^{2+}}=0.529\left[\frac{1^2}{3}\right]\stackrel{\circ }{A}$
For $B{e}^{3+}$ atom: $n=1,Z=4$
$r_{B{e}^{3+}}=0.529\left[\frac{1^2}{4}\right]\stackrel{\circ }{A}$
Upon taking ratio, $r_{H{e}^{+}}:r_{L{i}^{2+}}:r_{B{e}^{3+}}=\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$

On simplifying, we get the ratio as: $6:4:3$