Question

If the radius of $3^{rd}$ Bohr’s orbit of H-atom is r, then the radius of $4^{th}$ Bohr’s orbit of $H{e}^{+}$ would be

• A. $r\times \frac{4}{3}$
• B. $r\times \frac{16}{9}$
• C. $r\times \frac{8}{9}$
• D. $r\times \frac{9}{8}$

Answer 1

The correct option is C $r\times \frac{8}{9}$

The Bohr's radii are given by the expression, $r_n=\frac{n^2{r}_0}{Z}$
From the question,
$r_3=r=\frac{3^2{r}_0}{1}$
$\Rightarrow r_0=\frac{r}{9}$
Bohr's theory can be applied for hydrogen like atom. So,
$r_4\left(H{e}^{+}\right)=\frac{4^2{r}_0}{2}=\frac{8r}{9}$