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An a-particle of energy 5 MeV is scattered through 180dgree by a fixed uranium nucleus. The distance of the closest approach is of the order of

An a-particle of energy 5 MeV is scattered through 180dgree
by a fixed uranium nucleus. The
distance of the closest approach is of the order of

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
576-1768_we.PNG
Vikas TU
14149 Points
3 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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