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In a head on collision between an alpha ...

In a head on collision between an alpha particle and gold nucleus, the minimum distance of approach is `4xx10^(-14)m`. Calculate the energy of of `alpha`-particle. Take Z=79 for gold.

A

`7xx10^(-6)`J

B

`9.1xx10^(-18)` J

C

`8xx 10^(17)` J

D

`4xx10^(-5)`J

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The correct Answer is:
B
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