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What is the distance of closest approach...

What is the distance of closest approach when a `5.0 MeV` proton approaches a gold nucleus?

A

`4xx10^(-14)m`

B

`2X10^(-14)m`

C

`3xx10^(-14)m`

D

None of these

Text Solution

Verified by Experts

`E=((Ze)(e))/(4piepsilon_(0)r)=(9xx10^(9)xx79xx(1.6xx10^(-19))^(2))/(5xx1.6xx10^(-13))`
`r=2.3xx10^(-14)m`
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