The radius of hydrogen atom in its ground state is $5.3 \times 10^{-11}$ m. After collision with an electron it is found to have a radius of $21.2 \times 10^{-11}$ m. What is the principal quantum number n of the final state of the atom.
Answer & explanation
Correct answer: option 2
$r \propto n^2 \text { i.e. } \frac{r_f}{r_i}=\left(\frac{n_f}{n_i}\right)^2$
$\Rightarrow \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=\left(\frac{n}{1}\right)^2 \Rightarrow n^2=4 \Rightarrow n=2$