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 key 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`-particle momen-tarily 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 a-particle is at its stopping point. Then we can write the conservation of energy `E_(i) = E_(f)` as
K = `(1)/(4 pi epsilon_(0)) ((2e)(Ze))/(e) = (2Ze^(2))/(4 pi epsilon_(0)d)`
Thus the distance of closest approach dis given by
d = `(2Ze^(2))/(4 pi epsilon_(0)K)`
The maximum kinetic energy found in `alpha`-particles of natural origin is 7.7 MeV or `1.2 xx 10^(-12)` J.
d = `((2) (9.0 xx 10^(9) Nm^(2) //C^(2) ) (1.6 xx 1)^(-19) C^(2) Z)/(1.2 xx 10^(-12))`
= 3.84 `xx 10^(-16) = 3.03 xx 10^(-14)` = 30 fm
The atomic number of foil material gold is Z = 79, so that d(Au) = `3.0 xx 10^(-14)` m = 60 fm.
(1 fm (i.e. fermil) `= 10^(-15)`m )
The radius of gold nucleus is, therefore, less than `3.0 xx 10^(-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 that 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.