The radius of the hydrogen atom in its ground state is m. After collision with an electron it is fou — Atomic Structure Chemistry Question
Question
The radius of the 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. The principal quantum number of final state of the atom is
Answer: A
💡 Solution & Explanation
Radius of an orbit $r_n = r_1 \times n^2$. We are given $21.2 \times 10^{-11} = 5.3 \times 10^{-11} \times n^2 \implies n^2 = \frac{21.2}{5.3} = 4 \implies n = 2$. Therefore, correct answer is A.
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