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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Feb 2008 11:57:18 IST
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alpha particle of energy 5Mev is scattered through 180 degrees by a fixed uranium nucleus. the distance of closest approach?
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3 Feb 2008 01:45:39 IST
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Simply use energy conservation.
KE = PE
5 MeV = 2ke(92e)/r^2
Solve for r. Thats the ans
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