Question

An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point change. The atomic number of gold is 79, and the alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons).

Answer 1

From the conservation of energy, the kinetic energy of the alpha particle must be equal to the electrostatic potential energy between the alpha particle and gold nucleus at the distance of closest approach. The kinetic energy of the alpha particle is \(10.0 \mathrm{MeV}\). \(K=10.0 \mathrm{MeV}\left(\frac{10^{6} \mathrm{eV}}{1 \mathrm{MeV}}\right)\left(\frac{1.6 \times 10^{-19} \mathrm{~J}}{1 \mathrm{eV}}\right)\)

\(=1.6 \times 10^{-12} \mathrm{~J}\)

The electrostatic potential energy between the alpha particle and gold nucleus is, \(U=k \frac{q_{1} q_{2}}{d}\)

\(U=k \frac{(2 e)(79 e)}{d}\)

\(U=158 k \frac{e^{2}}{d}\)

\(U=158\left(9 \times 10^{9}\right) \frac{\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{d}\)

\(U=\frac{3.64032 \times 10^{-26} \mathrm{~J}}{d}\)

From the conservation of energy,

\(U=K\)

\(\frac{3.64032 \times 10^{-26} \mathrm{~J}}{d}=1.6 \times 10^{-12} \mathrm{~J}\)

\(d=2.275 \times 10^{-14} \mathrm{~m}\) or \(22.75 \mathrm{fm}\)