Question

What is the probability of an alpha particle getting ionized by gold electrons? What would happen next?​

Answer 1

Answer:

Down vote

The alpha particles used in this experiment have an energy of about 5.59 MeV. This energy is too high; the alpha particle cannot simply capture two electrons and form a stable helium atom. By way of comparison, the first and second ionization energy of helium is 24.6 eV and 54.4 eV, respectively.

Nevertheless, alpha particles cause frequent direct ionizations within a narrow diameter around a relatively straight track. This transfer of energy from the alpha particles to electrons causes an approximately continuous deceleration of the alpha particle. For 5.59 MeV alpha particles in gold, the initial electronic stopping power is about 220 MeV cm2 g−1. With a high density of 19.3 g cm−3, this means the initial linear energy transfer is about 4200 MeV cm−1. Therefore, the alpha particle loses its energy very quickly so that the range of the alpha particle in gold is very short. The continuous slowing down approximation (CSDA) range of 5.59 MeV alpha particles in gold is about 0.020 g cm−2. Considering the density of 19.3 g cm−3 again, the resulting range in gold is about 0.0010 cm. (Clearly, the gold foil in the experiment had to be very thin.)

Only after the alpha particle has slowed down like this, it can finally capture two electrons and form a stable helium atom.