Question

In a Beryllium atom, if $a_0$ be the radius of the first orbit, then the radius of the $n^{\text{th}}$ orbit will be,

• A. $n{a}_0$
• B. $a_0$
• C. $n^2{a}_0$
• D. $\frac{a_0}{n^2}$

Answer 1

The correct option is C $n^2{a}_0$

Radius of an electron in $n^{\text{th}}$ Bohr's orbit is given as,
$r_n=0.529\times \frac{n^2}{Z}\mathring{\text{A}}$
$\Rightarrow r_n\propto n^2$
Let $r_1$and $r_2$ be the radii of first and second orbit respectively, then,

$r_1\propto n_1^2-----\left(1\right)$
$r_2\propto n_2^2-----\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$, we get,
$\frac{r_1}{r_2}=\frac{n_1^2}{n_2^2}$

$\Rightarrow \frac{r_1}{r_2}=\frac{1}{4}$

$\Rightarrow r_2=4r_1=4a_0$
So in general we can say that,
$r_2=n^2{a}_0$
Where $n=$ orbit state, it can be $1,2,\mathrm{3......}$

Hence, $\left(C\right)$ is the correct answer.

Why this question ? This question is based on the concept of radius of $n^{\text{th}}$ orbit of Bohr's hydrogen atom.