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 1.00e10-15 m of the target? Assume that there is a head-on collision and that the target is fixed in place
To determine the kinetic energy required to bring the moving proton within a distance of 1.00e-15 m of the stationary proton, we can use the principles of energy conservation.
At a distance of 1.00e-15 m, the electrostatic potential energy between the two protons will convert entirely into kinetic energy. The electrostatic potential energy is given by the formula:
U = k * (q1 * q2) / r
Where:
U is the electrostatic potential energy
k is the Coulomb constant (8.99e9 Nm^2/C^2)
q1 and q2 are the charges of the two protons (both +1.60e-19 C)
r is the distance between them (1.00e-15 m)
Since the target proton is stationary, the initial kinetic energy of the moving proton will be entirely converted into potential energy as it approaches the target.
At the point of closest approach, all of the initial kinetic energy will be converted into potential energy. Therefore, the kinetic energy required to bring the proton within 1.00e-15 m of the target is equal to the initial kinetic energy.
The expression for kinetic energy is:
K = 0.5 * m * v^2
Where:
K is the kinetic energy
m is the mass of the proton (1.67e-27 kg)
v is the velocity of the proton
Since the target is fixed in place and there is a head-on collision, the direction of the velocity of the moving proton will change. Therefore, we need to consider the change in velocity in order to calculate the kinetic energy.
Using the principle of conservation of linear momentum, we can equate the initial and final momenta of the system:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Since the second proton is fixed in place, its final velocity is zero:
m1 * v1_initial = m1 * v1_final
Therefore, the magnitude of the initial velocity of the moving proton is equal to the magnitude of its final velocity:
|v1_initial| = |v1_final|
Now, we can combine the equations for potential energy and kinetic energy:
0.5 * m * |v1_initial|^2 = k * (q1 * q2) / r
Solving for |v1_initial|, we get:
|v1_initial| = sqrt((2 * k * (q1 * q2)) / (m * r))
Substituting the given values into the equation, we can calculate the required kinetic energy:
|v1_initial| = sqrt((2 * 8.99e9 Nm^2/C^2 * (1.60e-19 C * 1.60e-19 C)) / (1.67e-27 kg * 1.00e-15 m))
|v1_initial| ≈ 1.03e7 m/s
Finally, we can calculate the kinetic energy using the equation:
K = 0.5 * m * |v1_initial|^2
Substituting the values, we find:
K ≈ 0.5 * 1.67e-27 kg * (1.03e7 m/s)^2 ≈ 8.62e-11 J
Therefore, the kinetic energy required to bring the moving proton within 1.00e-15 m of the target is approximately 8.62e-11 Joules.