A beam of -particles is projected with the same kinetic energy towards four different metal foils: C — Atomic Structure Chemistry Question
Question
A beam of $\alpha$ -particles is projected with the same kinetic energy towards four different metal foils: Cu ($Z=29$), Ag ($Z=47$), Au ($Z=79$), and Ca ($Z=20$). For which metal is the distance of closest approach the minimum? Enter the atomic number $Z$ of that metal.
Answer: 20
💡 Solution & Explanation
The distance of closest approach is directly proportional to the charge of the target nucleus, $Z$. $r_{min} = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K.E.}$. Minimum distance is achieved when $Z$ is minimum. Among 29, 47, 79, and 20, Calcium ($Z=20$) is the smallest. [53, 54]
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