An a-particle of energy 5 MeV is scattered through 180dgreeby a fixed uranium nucleus. Thedistance of the closest approach is of the order of
sudhanshu , 11 Years ago
Grade 12
2 Answers
Saurabh Koranglekar
Last Activity: 4 Years ago
Vikas TU
Last Activity: 4 Years ago
One point is (_(92)235U) nranium nucleus ∴Q1=92e The other point charge is a particle ∴Q2=+2e Here the loss in K.E.=Ga∈∈P.E.(tilla−partic≤sthedistanced) ⇒1/2mν2=kq1q2/r⇒r=k2q1q2/(1/2mν2) ∴r=(9×10^9×2×1.6×10^−19×92×1.6×10^−19)/5×1.6×10^−13 =529.92×10^−16m =592.92×10^−14cm=5.2992×10^−12cm
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