Homework Help: Physics

Posted by Sui on Sunday, October 2, 2011 at 4:40pm.

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?

Physics - Sui, Sunday, October 2, 2011 at 4:41pm
Btw I multiplyed 2.31e-13 with 6.24e18 to get the final answer in MeV.
Physics - bobpursley, Sunday, October 2, 2011 at 4:53pm
I didn't check calcs, but it it the correct method.

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