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Question

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
1010cm
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C
1012cm
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D
1015cm
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Solution

The correct option is C 1012cm
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

5×106×1.6×1019 J=9×109×2×1.6×1019×92×1.6×10191r0=5.3×1014 mr0=5.3×1012 cm Hence, radius of closest approach is of the order of 1012 cm . Hence, option (c) is correct.

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