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 a-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 a-particle momentarily stops. The initial energy `E_i` is just the kinetic energy K of the incoming a- 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.
Let d be the centre-to- centre distance between the a-particle and the gold nucleus when the a- article 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))/(d) = (2 Ze^2)/(4 pi epsilon_0 d)`
Thus the distance of closest approach d is given by
`d = (2 Z e^2)/(4 pi epsilon_0 K)`
The miximum kinetic energy found in `alpha`- particles of natural origin is `7.7 MeV` or `1.2 xx 10^(-12) J`. Since `1//4 pi epsilon_0 = 9.0 xx 10^(9) N m^2//C^2`. Therefore with `e = 1.6 xx 10^(-19) C`, we have
`d = ((2)(9.0 xx 10^(9) Nm^2//C^2)(1.6 xx 10^(-10)C)^2 Z)/(1.2 xx 10^(-12) J)`
`= 3.84 xx 10^(-16) Z m`.
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 a-particle. Thus, the `alpha`-particle reverses its motion without ever actually touching the gold nucleus.