I knowing what i'm looking for -- alpha particle colliding with lead nucleus

[Eric Peraza]

Homework Statement

An alpha particle with kinetic energy 15.0MeVmakes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.30×10−12m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

What is the distance of closest approach?

Homework Equations

What is going on? This is an E&M class and this doesn't seem to have to do with potentials.
any help on where to start or anything would be great.

 

[lightgrav]

wow, thorough expectations here.
Is this angular momentum conserved? if so, what L does it have at closest approach?
So, how much KE does it have at closest approach? (no, L²/2mR² is not zero)
hmm ... where did the REST of the KE go?

 

[Eric Peraza]

what does it mean by closest approach? that's where i mainly got thrown off. and do i calculate the change in voltage and then calculate the change in energy? I am not sure what the question is asking by "closest approach" but my guess is something to do with voltage and energy

 

[Dick]

what does it mean by closest approach? that's where i mainly got thrown off. and do i calculate the change in voltage and then calculate the change in energy? I am not sure what the question is asking by "closest approach" but my guess is something to do with voltage and energy

It's a mechanics problem as well as an EM problem. Since you are assuming the lead nucleus is stationary, the angular momentum of the alpha particle is conserved. At closest approach the velocity of the particle is perpendicular to the displacement vector connecting it with the nucleus. Matching that with the given initial angular momentum gives you a relation between v and r. Considering the energy lost by the particle due to the potential gives you another. Solve them simultaneously.

 

[PhysyCola]

Hey there Eric. If you are still having trouble with this problem, you could consider the total energy of the system. In polar coordinates (where r is the distance of the alpha particle from the lead nucleus), the kinetic energy of the system can be written as, noting that the mass is relativistic (as the alpha particle has 15MeV of energy):

KE = 1/2*m*r(dot)^2 + J^2/(2*m*r^2)

As we can treat this as a central force problem, the potential is given by:

U = -(k*q1*q2)/r

Summing these two together gives:

E = 1/2*m*r(dot)^2 + J^2/(2*m*r^2) - (k*q1*q2)/r

Now, what do we know about r at the distance of closest approach? How can we use this to find r?

Let me know if you are still stuck at this point, but I hope that helps.

PS. Sorry for the lack of latex formatting, I am still learning it!

 

[lightgrav]

15 MeV would be relativistic for an electron, but not for a proton or alpha.
you can use regular Energy conservation (quadratic)