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 $212\ \times \ 10^{- 11}$ m. What is the principal quantum number $n\ $of the final state of atom?
A) $n = 4$
B) $n = 2$
C) $n = 16$
D) $n = 3$
Explanation
(b) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r \propto n^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{f}}{r_{i}} = \left( \frac{n_{f}}{n_{i}} \right)^{2}$
$\ \ \ \ \ \ \ \ \ \frac{21.2\ \times 10^{- 11}}{5.3 \times 10^{- 11}}$ = $\left( \frac{n}{1} \right)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n^{2} = 4$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.