ElectrostaticsHard
Question
An alpha particle of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The distance of closest approach is of the order :
Options
A.1(Ao)
B.10-10 cm
C.10-12 cm
D.10-15 cm
mv2 = 
⇒ d = 10-12 cmMore Electrostatics Questions
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