In a beryllium atom, if a0 be the radius of the first orbit, then the radius of the second orbit will be in general
Answer & explanation
Correct answer: option 3
$r \propto n^2 \quad \Rightarrow \quad r_n=n^2 a_0 \quad\left(∵ r_1=a_0\right)$
In a beryllium atom, if a0 be the radius of the first orbit, then the radius of the second orbit will be in general
Correct answer: option 3
$r \propto n^2 \quad \Rightarrow \quad r_n=n^2 a_0 \quad\left(∵ r_1=a_0\right)$