Question

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. If the energy of the doubly charged alpha nucleus was (5.00 MeV), how close to the gold nucleus (79 protons) could it come before being deflected?

a) (2.64 X 10^14 m)
b) (3.88 X 10^14 m)
c) (4.72 X 10^14 m)
d) (6.19 X 10^14 m)

Answer 1

The alpha particle can come as close as 4.72 x 10^14 m before being deflected by the gold nucleus.

Step-by-step explanation:

The path of an alpha particle can be deflected by the Coulomb interaction when it approaches a gold nucleus. In this experiment, the energy of the alpha particle was 5.00 MeV. To determine how close the alpha particle could come before being deflected, we need to consider the energy conversion from kinetic energy to potential energy. The potential energy of the alpha particle near the gold nucleus can be calculated using the equation:

U = (q_1 * q_2 * k) / r, where U is the potential energy, q_1 and q_2 are the charges of the alpha particle and gold nucleus (in this case, +2e and +79e, respectively), k is the Coulomb constant, and r is the distance between the particles.

Using the given energy of 5.00 MeV and the conversion factor 1 MeV = 1.6 * 10^-13 J, we can calculate the potential energy of the alpha particle. Equating this potential energy to the kinetic energy, we can solve for the distance, r, between the alpha particle and the gold nucleus.

After performing the calculations, the correct answer is (c) 4.72 x 10^14 m.