A beam of -particles is projected with the same kinetic energy towards four different metal foils: Cu (), Ag (), Au (), and Ca (). For which metal is the distance of closest approach the minimum? Ente

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.

Prashant Kotian · Accepted Answer

Correct answer: 20. The distance of closest approach is directly proportional to the charge of the target nucleus, . . Minimum distance is achieved when is minimum. Among 29, 47, 79, and 20, Calcium () is the smallest. [53, 54]

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