In a head on collision between an alpha particle and gold nucleus (Z=79), the distance of closest approach is 39.5 fermi. Calculate the energy of `alpha`-particle.

Explanation

Given `r_(0)=39.5 "fm"=39.5xx10^(-15)m, =79, E=?`
As `E=(Ze(2e))/(4 pi epsilon_(0)r)=(2 Ze^(2))/(4 pi epsilon_(0)r)`
`=(9xx10^(9)xx79xx2xx(1.6xx10^(-19))^(2))/(39.5xx10^(-15))`
`=(9xx79xx2xx1.6xx1.6xx10^(-14))/(39.5)J`
or `E=(18xx79xx1.6xx1.6xx10^(-14))/(39.5xx1.6xx10^(-13))`
`=5.76` M eV.