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Question

An α -particle of energy 5 MeV is scattered through 180 by gold nucleus. The instance of closest approach is of the order of



A

1014cm

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B

1010cm

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C

1016cm

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D

1012cm

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Solution

The correct option is D

1012cm


Distance of closest approach is = 1014m=1012cm (Order of size of nucleus)


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