The charge of the gold nucleus is Ze where Z is the atomic number of gold. α-particles are nuclei of a helium atom, thus carrying 2 units (2e). Thus the calculation was needed for a single nucleus. Scattering of the α-particlesĪs we all know, the gold foil is very thin, so we can assume that the α-particles will not suffer more than one collision. Thus, because of such a large empty space in an atom, it is evident that most of the α-particles pass right through the atom and when any α-particles come near the nucleus, they are scattered due to the electric field. Rutherford said the size of the atom will be 10 -15 approximately but in actual it was 10 -10 and as electrons move around at some distance the atom is mostly empty space. So, in Rutherford’s model of the atom, all the positive charge and the mass are at the centre and the electrons will move around it like planets around the sun at some distance. Rutherford said from the observation that the greater repulsive force can be provided if the mass and positive charge is concentrated at the center which will cause the reflection. This means that the particle was scattered completely to the backward direction which means there must be some kind of big repulsive force This experiment was conducted and the result was that in the scattered beam of light, many particles pass through the foil without collusion, just 0.14% of incident particles scatter more than 1 o and 1 in 8000 particles deflect more than 90 o. α-particles emitted by the source are concentrated into a single beam of light which was allowed to strike the thing gold foil and the concentrated beam is scattered after hitting and can be seen in the screen through a microscope. They used a beam of 5.5 MeV emitted from a radioactive source (Bismuth) at a thin layer of foil made of gold.
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