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.
Ok, so I'm thinking to use:
KE=U
Ke=-qV=-q*k*q/r
=-1.602e-19*8.99e9*1.602e-19/1.00e-15
= 2.31e-13
=1.44MeV
Does that make sense? Am I doing something wrong?
Btw I multiplyed 2.31e-13 with 6.24e18 to get the final answer in MeV.
I didn't check calcs, but it it the correct method.
Your approach is correct, but there is a calculation error. Let's go through it step by step:
To calculate the kinetic energy required for the proton to come within 1.00e-15 m of the stationary proton, you can consider the electrostatic potential energy being converted into kinetic energy.
The electrostatic potential energy (U) between two protons is given by:
U = (k * q1 * q2) / r,
where k is the Coulomb constant (8.99e9 N*m^2/C^2), q1 and q2 are the charges of the protons (both +1.602e-19 C since they are protons), and r is the separation distance (1.00e-15 m in this case).
Therefore, we can rewrite the equation as:
U = (8.99e9 * (1.602e-19)^2) / 1.00e-15.
Evaluating this expression, we get:
U ≈ 2.31e-13 J.
Now, to find the kinetic energy (KE) needed for the proton to approach the stationary proton, we can equate it to the potential energy (U):
KE = U = 2.31e-13 J.
Finally, we can convert this energy into MeV using the conversion factor 1 MeV = 1.602e-13 J:
KE ≈ (2.31e-13 J) / (1.602e-13 J/MeV) ≈ 1.44 MeV.
Therefore, the correct answer is approximately 1.44 MeV.