# Problem: In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated towards a gold nucleus and its path was substantially deflected by the Coulomb interaction. If the initial kinetic energy of the doubly charged alpha nucleus was 5.10 MeV, how close to the gold nucleus (79 protons) could it come before being turned around?____________m

###### FREE Expert Solution

We get away with classical mechanics because the speed is low.

Using conservation of energy:

$\overline{){\mathbf{P}}{{\mathbf{E}}}_{{\mathbf{i}}}{\mathbf{+}}{\mathbf{K}}{{\mathbf{E}}}_{{\mathbf{i}}}{\mathbf{=}}{\mathbf{P}}{{\mathbf{E}}}_{{\mathbf{f}}}{\mathbf{+}}{\mathbf{K}}{{\mathbf{E}}}_{{\mathbf{f}}}}$

The energy for the alpha particles will be repelled by the nucleus. Therefore, at its closest point, all the KE transforms to PE.

PE = k (q1q2/r) ###### Problem Details

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated towards a gold nucleus and its path was substantially deflected by the Coulomb interaction. If the initial kinetic energy of the doubly charged alpha nucleus was 5.10 MeV, how close to the gold nucleus (79 protons) could it come before being turned around?

____________m