Question

The difference between $(n+2{)}^{th}$ Bohr radius and $n^{th}$ Bohr radius is equal to the $(n-2{)}^{th}$ Bohr radius. The value of $n$ is:

Answer 1

$r=\frac{n^2{h}^2{ϵ}_0}{Z\pi m{e}^2}=0\cdot 529\times \frac{n^2}{Z}{A}^0$

As $r_n\propto n^2$ so, $r_{n+2}=k(n+2{)}^2;r_n=k{n}^2$ and $r_{n-2}=k(n-2{)}^2$

Here, $k(n+2{)}^2-k{n}^2=k(n-2{)}^2$
$n^2+4n+4-n^2=n^2-4n+4$
$n^2=8n\Rightarrow n=8$