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Question
Determine the distance of the closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a nucleus of Z = 80, stops and reverses its direction.
Solution
At the distance of nearest approach
PE = KE
`("k"("ze")(2"e"))/("r"_0) = 4.5 "Me" "V" = 4.5 xx 10^6 xx 1.6 xx 10^-19 "J"`
`"r"_0 = ("k"("ze")(2"e"))/(4.5 xx 1.6 xx 10^-13)`
= `(9 xx 10^9 xx (80) xx 2 xx (1.6 xx 10^-19)^2)/(4.5 xx 1.6 xx 10^-13) = 51.2 xx 10^-15 "m"`.
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