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The distance of closest approach of 1,00...

The distance of closest approach of 1,00 MeV protons incident on gold (Z=79) nuclei will be:

A

`1.14 xx 10 ^(-15) m`

B

`6.28 xx 10 ^(-15) m`

C

`1.44 xx 10 ^(-15) m`

D

`8.99 xx 10 ^(-15) m `

Text Solution

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The correct Answer is:
A
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