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An alpha-particle having energy = 10 meV...

An `alpha`-particle having energy = 10 meV collides with a nucleus of atomic number 50. The distance of the closest approach is

A

`1.5xx10^(-16) m`

B

`1.5 xx10^(-19)`

C

`1.5xx10^(-7) m`

D

`1.5 xx 10^(-12)` m

Text Solution

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The correct Answer is:
C
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