An alpha particle with kinetic energy `10 Me V` is heading toward a stationary tin nuclcus of atomic number 50. Calculate the distance of closest approach (Fig . 3.23).
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Answer

A, D

Explanation

Due to repulsion by the nucleus, the kinetic enerty of the `alpha`-particle gradually decreases at the expense of electrostatic potential energy.
:. Decrease in kinetic energy= increase in potential energy
or `1/2 mv^2=U_f-U_i`
or `1/2 mv^2=1/(4piepsilon_0).(q_1q_2)/r-0`
`: r=1/(4piepsilon_0).((2e)(50e))/(KE)`
Substituting the values, `r=((9xx10^9)(2xx1.6xx10^(-19))(1.6xx10^(-19)xx50))/(10xx10^6xx1.6xx10^(-19))`
`=14.4xx10^-15m`