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 momentarily stops. The initial energy `E_(i)` is just the kinetic energy K of the incoming - particle. The final energy `E_(f)` is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. (12.1).
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)/(4piepsilon_(0))((2e)(Ze))/(d)=(2Ze^(2))/(4piepsilon_(0)d)`
Thus the distance of closest approach d is given by
`d=(2Ze^(2))/(4piepsilon_(0)K)`
The maximum kinetic energy found in `alpha`-particles of natural origin is 7.7 MeV or `1.2 xx 10^(12)`J. Since `1//4piepsilon_(0)=9.0xx10^(9)Nm^(2)//C^(2)`. Therefore with e = `1.6 xx 10^(-19)C`, we have,
`d=((2)(9.0xx10^(9)Nm^(2)//C^(2))(1.6xx10^(-19)C)^(2)Z)/(1.2 xx 10^(-12)J)`
`=3.84 xx 10^(-16) Zm`
The atomic number of foil material gold is Z = 79, so that d (Au) = `3.0 xx 10^(-14)m` = 30 fm. (1 fm (i.e. fermi) = `10^(-15)` m.)
The radius of gold nucleus is, therefore, less than `3.0 × 10^(–14)` m. 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.
