Q. 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
A) $n = 4$
B) $n = 2$
C) $n = 16$
D) $n = 3$
Explanation
(b) $r \propto n^{2},i.e.,\frac{r_{f}}{r_{i}} = \left( \frac{n_{f}}{n_{i}} \right)^{2}$
$\Rightarrow \frac{21.2 \times 10^{- 11}}{5.3 \times 10^{- 11}} = \left( \frac{n}{1} \right)^{2} \Rightarrow n^{2} = 4 \Rightarrow n = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.