Question

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

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

Answer 1

The correct option is B $6:3:2$

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 $H$ atom: $n=1,Z=1$
$r_H=0.529\left[\frac{1^2}{1}\right]\stackrel{\circ }{A}$

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: $n=1,Z=3$
$r_{L{i}^{2+}}=0.529\left[\frac{1^2}{3}\right]\stackrel{\circ }{A}$

Upon taking ratio, $r_H:r_{H{e}^{+}}:r_{L{i}^{2+}}=1:\frac{1}{2}:\frac{1}{3}$

On simplifying, we get the ratio as:
$6:\frac{6}{2}:\frac{6}{3}i.e.,6:3:2$