# Alpha Particle shot at Nucleus Energy

1. Feb 4, 2009

### pearbear21

1. The problem statement, all variables and given/known data
What energy alpha particle would be needed to just reach the surface of an Al nucleus if its radius is 4fm?

*This problem is found in a section dealing with Rutherford's gold foil scattering experiment, so take it in that context.

2. Relevant equations
F=(kqQ)/r2

3. The attempt at a solution
I assume the book (Modern Physics by Tipler and Llewellyn ed. 5) wants this performed a classical manner, so I started by considering the energy of the particle as an integral of the coulombic force equation from r=$$\infty$$ to r=4X10-15m.

so, kqQ$$\int$$r-2dr=kqQ[-r-1]

evaluated from $$\infty$$ to 4X10-15m gives a value of -1.4995992X10-12 Joules. This seems quite unreasonable as the repulsion seems intuitively large at such small distances and a joule is a small unit to begin with.

What do you think of my method and answer? Do you feel there is a different way I should have approached the problem? At first, I tried finding the force to hold the alpha particle just on the surface of the Al nucleus, but after finding that force couldn't find a way to convert to energy correctly.

2. Feb 4, 2009

### Brian_C

The energy you are trying to calculate is the final electrostatic energy of the system. When the alpha particle is fired with zero impact parameter, the initial kinetic energy of the alpha particle is entirely converted into electrostatic energy at the point of closest approach. Just use conservation of energy.

3. Feb 5, 2009

### pearbear21

Thanks, I think I figured it out upon further inspection of the text. rd, distance of closest approach, may be calculated as follows:

rd = kqQ/.5mv2