In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV a-particle before it comes momentarily to rest and reverses its direction?

Explanation

Kinetic energy of `alpha`- particle K=7.1 MeV
`1eV=1.6xx10^(-19)J, (1)/(4pi epsi_(0))=k=9xx10^(9)Nm^(-2)C^(-2)`
`e=1.6xx10^(-19)` charge on `alpha`- particle =2
Atomic number of nucleus of gold Z = 79
`rArr` During the process of scattering the total mechanical energy of the `alpha`-particle and the system that is made of the nucleus of gold remains constant.
Suppose, the system.s initial mechanical energy is `E_(i)` before the particle and nucleus interact and it is equal to its mechanical energy `E_(f)` when the `alpha`-particle stops. Here `E_(i)=` kinetic energy K and `E_(f)` = electric potential energy U.
Let d be the centre to centre distance between the `alpha`-particle and the gold nucleus when the `alpha`-particle is at its stopping point.
In potential energy of system. `U=(kq_(1)q_(2))/(r)`
`q_(1)=2e, q_(2)=Ze`
`:.U=(K(2e)(Ze))/(d)`
Energy is conserved.
`:.E_(i)=E_(f)`
`:.K=U`
`:.K=(2kZe^(2))/(d)`
`:.d=(2kZe^(2))/(K)`
`:.d=(2xx9xx10^(9)xx79xx(1.6xx10^(-19))^(2))/(7.7xx10^(6)xx1.6xx10^(-18))`
`=295.48xx10^(-16)m`
`:.d=30xx10^(-15)m=30 f_(m)[1f_(m)=10^(-15)m]`