Question

Nuclear fusion reactions require that positively charged nuclei be brought into close proximity, against the electrostatic repulsion. As a simple example, suppose a proton is fired at a second, stationary proton from a large distance away. What kinetic energy must be given to the moving proton to get it to come within \(1.00 \cdot 10^{-15} \mathrm{~m}\) of the target? Assume that there is a head-on collision and that the target is fixed in place.

Short answer

Answer: The initial kinetic energy required for the proton is \(2.30 \cdot 10^{-13} \mathrm{~J}\).

Step-by-step solution

Identify the knowns and what we need to find

We know the distance that the protons need to come within, as well as the electric charge on a proton. We need to find the initial kinetic energy required for the moving proton.

Calculate the electric potential energy

When the protons are at the closest separation of \(1.00 \cdot 10^{-15} \mathrm{~m}\), the electric potential energy can be calculated using Coulomb's Law: \(U = \frac{k \cdot q_1 \cdot q_2}{r}\) where \(U\) is the potential energy, \(k = 8.99 \cdot 10^9 \mathrm{~N m^2/C^2}\) is the electrostatic constant, \(q_1 = q_2 = 1.60 \cdot 10^{-19} \mathrm{~C}\) are the charges of the protons, and \(r = 1.00 \cdot 10^{-15} \mathrm{~m}\) is the distance between the protons. Calculate the energy: \(U = \frac{(8.99 \cdot 10^9) \cdot (1.60 \cdot 10^{-19})^2}{1.00 \cdot 10^{-15}}\)

Solve for the electric potential energy

Performing the calculation, we get: \(U \approx 2.30 \cdot 10^{-13} \mathrm{~J}\) So, the electric potential energy when the two protons are \(1.00 \cdot 10^{-15} \mathrm{~m}\) apart is \(2.30 \cdot 10^{-13} \mathrm{~J}\).

Use energy conservation to find the initial kinetic energy

Since energy is conserved, the initial kinetic energy of the moving proton, \(K\), must be equal to the sum of the final kinetic energy and the final electric potential energy. However, the final kinetic energy of the moving proton is assumed to be zero, so the initial kinetic energy is equal to the electric potential energy: \(K = U = 2.30 \cdot 10^{-13} \mathrm{~J}\)

Write the final answer

The initial kinetic energy required for the moving proton to come within \(1.00 \cdot 10^{-15} \mathrm{~m}\) of the stationary proton is \(2.30 \cdot 10^{-13} \mathrm{~J}\).