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Question

An αparticle of energy 5 MeV is scattered through 180 by gold nucleus. The distance of closest approach is of the order of

A
1012 cm
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B
1016 cm
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C
1010 cm
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D
1014 cm
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Solution

The correct option is A 1012 cm
Assume α particle is thrwon (head- on) from very far
Applying conservation of mechanical energy we can write
Kα+Vα=kf+Vf
5 MeV+0=0+9×109×79 e×2 er×1.6×1019×106MeV
r=9×109×19×2×1.6×1019106×5m
=4.55×1012cm

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