The radius of $2^{\mathrm{nd}}$ orbit of ${\mathrm{He}}^{+}$ of Bohr's model is $r_1$ and that of fourth orbit of ${\mathrm{Be}}^{3+}$ is represented as $r_2.$ Now the ratio $\frac{r_2}{r_1}$ is $x:1.$ The value of $x$ is _____.

The formula to calculate the radius $\left(r\right)$ of any $n^{\mathrm{th}}$ Bohr's orbit is given by

$r=r_0\frac{n^2}{Z} ...\left(1\right)$

where, $r_0$ is Bohr's radius, $n$ is the principal quantum number and $Z$ is the atomic number.

Thus, for Helium ion,

$\begin{array}{rcl}{r}_1& =& r_0\frac{2^2}{2}\\ & =& 2r_0 ...\left(2\right)\end{array}$

And, for the Beryllium ion,

$\begin{array}{rcl}{r}_2& =& r_0\frac{4^2}{4}\\ & =& 4r_0 ...\left(3\right)\end{array}$

Divide equation (3) by equation (2) to obtain the required ratio.

$\begin{array}{rcl}\frac{r_2}{r_1}& =& \frac{4r_0}{2r_0}\\ & =& 2\end{array}$

Hence, $x=2$.