Question

In a beryllium atom, if $a_0$ be the radius of the first orbit, then the radius of the second will be in general.

• 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 $n^2{a}_0$

$F\propto n^2$

$r=\frac{n^2{h}^2{ϵ}_0}{\pi mZ{e}^2}$

$r_1=\frac{h^2{ϵ}_0}{\pi mZ{e}^2}=a_0$

$r_n=\frac{n^2{h}^2{ϵ}_0}{\pi mZ{e}^2}$

$r_n=a_0{n}^2$