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What is the distance of closest approach of a 7.7 MeV alpha particle to a gold nucleus?
AAbout 6 fm
BAbout 30 fm
CAbout 300 fm
DAbout 0.8 fm
Answer & Solution
Correct answer: B. About 30 fm
Using d = 3.84 x 10^-16 Z m with Z = 79 gives about 3.0 x 10^-14 m, which is 30 fm.
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