Question

In a beryllium atom, if $a_0$be the radius of the first orbit, then the radius of the second orbit 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 $n^2{a}_0$

$r\alpha n^2\Rightarrow r_n=n^2{a}_0(\because r_1=a_0)$