An α- particle of energy 5MeV is scattered through 180o by a fixed uranium nucleus. The distance of the closest approach is of the order of:
A
1˚A
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B
10−10cm
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C
10−12cm
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D
10−15cm
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Solution
The correct option is C10−12cm
Given- ∗ Energy of α -particle =5MeV By conservation of energy, we can say that for the distance of closest approach all the kinetic energy is converted into potential energy. ⇒KE initial =PE final
⇒12mv2=kq1q2r,q1= charge of uranium nucleus q2= charge of α− particle
⇒5×106×1.6×10−19J=9×109×2×1.6×10−19×92×1.6×10−191⇒r0=5.3×10−14mr0=5.3×10−12cm Hence, radius of closest approach is of the order of 10−12cm . Hence, option (c) is correct.