Question

If the radius of second Bohr's orbit of hydrogen atom is $r_2$ then the radius of third Bohr's orbit of $B{e}^{3+}$ ion will be:

• A. $9r_2$
• B. $\frac{9r_2}{8}$
• C. $\frac{9r_2}{16}$
• D. $\frac{9r_2}{4}$

Answer 1

The correct option is C $\frac{9r_2}{16}$

Bohr's radius for $n^{th}$ orbit, $\left(r_n\right)=0.529\left[\frac{n^2}{Z}\right]\stackrel{\circ }{A}$
where $n=\text{Orbit number},Z=\text{Atomic number}$
$\left(r_n\right)=r_2$ for $2^{n}d$ orbit of hydrogen

Therefore, ratio of both the radius gives:
$\frac{r_H}{r_{B{e}^{3+}}}=\frac{n_H^2}{Z_H}\times \frac{Z_{B{e}^{3+}}}{n_{B{e}^{3+}}^2}=\frac{2^2\times 4}{1\times 3^2}$

$\Rightarrow \frac{r_2}{r_{B{e}^{3+}}}=\frac{16}{9}$

$\Rightarrow r_{B{e}^{3+}}=\frac{9r_2}{16}$