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An alpha particle of energy 5 MeV is sca...

An `alpha` particle of energy 5 MeV is scattered through `180^(@)` by a fixed uranium nucleus. The distance of the closest approach is of the order of ........

A

`1Å`

B

`10^(-10)cm`

C

`10^(-12)cm`

D

`10^(-15)cm`

Text Solution

Verified by Experts

The correct Answer is:
C
Since kinetic energy is converted into potentia energy
`:.` kinetic energy = potential energy
`:.5MeV=(1)/(4piepsi_(0)).(q_(1)q_(2))/(r)`
`:.5xx10^(6)xxe=(9xx10^(9)xx(92e)(2e))/(r)`
`:.r=(9xx10^(9)xx92xx2xxe)/(5xx10^(6))=5.3xx10^(-14)m`
`:.r=5.3xx10^(-12)cm`
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