As the positive particle approaches the positive nucleus it experiences a retarding force. The work done against this force is equal to the change in kinetic energy of the particle. In this case, the particle stops at the given distance from the nucleus, so all the Ke is used up and the particle velocity is zero at that distance. It gained potential energy of course and that will now push the particle away from the nucleus.
I suspect you have the equation for this gain in potential energy in your book but it is easy to derive.
F = k Q q/r^2 (Coulomb)
integral Work = F dr from oo to R (which is 35.9 fm)
---> k Q q (1/R) = Ke
Here's the question I'm currently working on...
What initial kinetic energy must an α particle have if it is to approach a stationary lead nucleus to within a distance of 35.9 fm?
So I looked up a couple of things I thought we need for this problem. I have the atomic radius of lead, and I know that the charge of the particle is 2*e, where e is the elementary charge. I was thinking of approaching this using Coulomb's law, but couldn't figure out how to get from Force to Kinetic Energy! HELP!
Thank you SO MUCH!
3 answers
so, q is the elementary charge times 2, and Q is the charge of lead? what is that?
and thanks, the concepts are much clearer.
and thanks, the concepts are much clearer.
The charge on a lead nucleus is the number of protons (the atomic number) times the absolute value of the charge on an electron.
1 answer