Atoms - Result Question 40
42. 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
[1994]
(a) $n=4$
(b) $n=2$
(c) $n=16$
(d) $n=3$
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Answer:
Correct Answer: 42. (b)
Solution:
- (b) $r \propto n^{2}$
$\therefore \frac{\text{ radius of final state }}{\text{ radius of initial state }}=n^{2}$
$\frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=n^{2}$
$\therefore n^{2}=4$ or $n=2$
