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 is \(2.30 \cdot 10^{-14} \mathrm{~J}\).

Step-by-step solution

Identify the required energy to overcome electrostatic repulsion

First, we need to find the electrostatic potential energy at a distance of \(1.00 \cdot 10^{-15} \mathrm{~m}\) between the protons. We will use the formula: \(U = k_e \frac{q_1q_2}{r}\). Here, \(k_e = 8.99 \cdot 10^9 \mathrm{~N~m^2/C^2}\) is the electrostatic constant, and \(q_1 = q_2 = 1.60 \cdot 10^{-19} \mathrm{~C}\) is the charge of the protons. The distance between charges, \(r = 1.00 \cdot 10^{-15} \mathrm{~m}\). Plug the values into the formula to find the required energy: \(U = 8.99 \cdot 10^9 \mathrm{~N~m^2/C^2} \cdot \frac{(1.60 \cdot 10^{-19} \mathrm{~C})^2}{1.00 \cdot 10^{-15} \mathrm{~m}}\). Now, calculate \(U\): \(U \approx 2.30 \cdot 10^{-14} \mathrm{~J}\). The energy required to overcome the electrostatic repulsion and get within the required distance is \(2.30 \cdot 10^{-14} \mathrm{~J}\).

Determine initial kinetic energy

The initial kinetic energy of the moving proton should be equal to the energy required to overcome the electrostatic repulsion (\(2.30 \cdot 10^{-14} \mathrm{~J}\)). So the initial kinetic energy is: \(K = 2.30 \cdot 10^{-14} \mathrm{~J}\) And hence, this is the kinetic energy that must be given to the moving proton to get it to come within \(1.00 \cdot 10^{-15} \mathrm{~m}\) of the stationary proton.