Elastic collision of an alpha particla and Lead

[panders3]

Homework Statement

An a particle of mass 6.6 x 10-27 kg came closest to an atom of lead with a distance of 4.8 x 10-14 m. At what velocity will the a particle rebound directly away if the collision is totally elastic?

Mass of the Alpha Particle - 6.6 x 10^-27 kg
Alpha particles are 2n2p

Mass of Lead Particle - 3.44 x 10^-25kg


Distance between the two particles when the elastic collision occurs - 4.8 x 10^-14 m

Homework Equations

v = f$$\lambda$$
Ek = hf – W

The Attempt at a Solution

I am not sure how to find the energy at which the elastic collision will occur. I assume it has to do with the atomic number of lead and the alpha particle, but i cannot find the theory behind it in my text.

 

[gneill]

Consider the potential energy due to the electric field that exists between the lead nucleus and the alpha particle at the instant of their minimum separation.

 

[panders3]

so then to find the electric field you use E= (kq1q2)/r where q1 would be the charge on the lead and q2 would be the charge on the alpha particle r would be the minimum distance. Then e would be the potential energy which would then be converted to kinetic energy as the alpha particle moves away.

I would need to assume that because the lead stays still?

 

[gneill]

Conveniently, the author of the problem did not specify a frame of reference to use. So if you were to consider the instant when the lead atom and the alpha particle are at their minimum distance (and at rest with respect to each other at that instant) and take the center of mass at that moment as the origin of your frame of reference, you may find your life much easier!