In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v `alpha`- particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .

Explanation

The idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an `alpha`-particle and a gold nucleus is conserved. 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`-particlemomentarily stops. The initial energy `E_(i)` is just the kinetic energy K of the incoming `alpha`-particle. The final energy `E_(f)` is just the electric potential energy U of the system.
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. Then we can write the conservation of energy `E_(i)=E_(f)` as
`K=(1)/(4pi epsilon_(0))((2e)(Ze))/(d)=(2Ze^(2))/(4pi epsilon_(0)d)`
Thus the distance of closest approach d is given by
`d=(2Ze^(2))/(4pi epsilon_(0)K)`
The maximum kinetic energy found in `alpha`-particles of natural origin is 7.7 MeV or `1.2xx10^(-12)` J.
Since `(1)/(4pi epsilon_(0))=9.0xx10^(9)Nm^(2)//C^(2),e=1.6xx10^(-19)C,`
we have, `d=((2)(9.0xx10^(9)Nm^(2)//C^(2))(1.6xx10^(-19)C)^(2)Z)/(1.2xx10^(-12))`
`=3.84xx10^(-16)=3.03xx10^(-14)=30fm`
The atomic number of foil material gold is Z = 79,
so that d(Au) `=3.0xx10^(-14)m=60` fm.
(l fm (i.e. fermil ) `=10^(-15)`m)
The radius of gold nucleus is, therefore, less than `3.0xx10^(-14)`. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is taht the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the `alpha`-particle. Thus, the `alpha`-particle reverses its motion without ever actually touching the gold nucleus.