In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z=75, when an `alpha` - particle of 5 Me V enery impinges on it before it comes momentarily to rest and reverses its direction.
How will the distance of closest approach be affected when the kinetic energy of the `alpha`-particle is doubled ?

Explanation

`K.E= 5 Me V. =75`
`"Potential energy "=(KZe^(2))/(d_(0))=(1)/(2) m^(1)v_(0)^(2)rArrd_(0)=(KZe^(2))/((M^(1)V_(0)^(2)/(2)))=(9xx10^(9)xx75xx2(1.6xx10^(-19))^(2))/(5xx1.6xx10^(-13))`
`=(3456xx10^(9)xx10^(-38))/(8xx10^(-13))=432xx10^(+22-38)=432xx10^(-16)m=43.2xx10^(-15)m=43.2 fm`
If, the kinetic energy is doubled the distance decreases (halved).