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Problem: In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged aplha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?

FREE Expert Solution

Electric potential energy for point charges:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{R}}}$

At the instance the alpha particle is deflected, its potential energy is equal to the initial energy.

U = 5.00 MeV(106eV/1MeV)(1.6 × 10-19J/eV) = 8.0 × 10-13

Qg = 79(1.6 × 10-19C) = 126.4 × 10-19 C

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Problem Details

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged aplha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?