In a Geiger- Marsden experiment, calculate the distance of closest approach to the nucleus of Z=80, when an `alpha` particle of 8 Me V energy 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

`Z=80, KE=8 Me V`
`"Potential energy "=(KZe^(2))/(d_(0))=(1)/(2)m^(1)v_(0)^(2)rArrd_(0)=(KZe^(2))/(((M'V_(0)^(2))/(2)))=(9xx10^(9)xx80xx(1.6xx10^(-19))^(2))/(8xx1.6xx10^(-13))" "[because 1 Me V=1.63xx10^(-13)J]`
`=(18xx80xx10^(9)xx1.6xx10^(-10))/(8xx10^(6))=128.8 fm`
`"Since "d_(0)alpha(1)/(E_(k))`
So, when kinetic energy is doubled the distance of closest to halved.
If, the kinetic energy is doubled the distance of `alpha` particles is halved.