Q 12-12-073JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
The radius of $2^\text{nd}$ orbit of $\text{He}^+$ of Bohr's model is $r_1$ and that of fourth orbit of $\text{Be}^{3+}$ is represented as $r_2$. Now the ratio $\dfrac{r_2}{r_1}$ is $x:1$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
$r\propto\dfrac{n^2}{Z}$: $r_1\propto\dfrac42=2$, $r_2\propto\dfrac{16}{4}=4$. $\dfrac{r_2}{r_1}=2$.
Solution by Sreeraj P, M.Sc Physics