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:

  1. (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 $n=2$