# Impact Parameter, Closest Approach

1. Nov 14, 2007

### genloz

1. The problem statement, all variables and given/known data
3. A 15MeV Alpha particle is incident at impact paramter b=0 on a Mg Nucleus, atomic number Z=12 and atomic mass A=24. What is the distance of closest approach?

2. Relevant equations
None given, but I thought the following might be relevant:
$$r_{min}=\frac{Zke^{2}}{KE}$$
$$b=\frac{Zke^{2}}{KE} * \sqrt{\frac{1+cos\theta}{1-cos\theta}}$$

3. The attempt at a solution
First I thought that given the above:
$$b=r * \sqrt{\frac{1+cos\theta}{1-cos\theta}}$$
$$0=r * \sqrt{\frac{1+cos\theta}{1-cos\theta}}$$
$$r=0$$

But then I thought perhaps the distance of closest approach is just the radius of the nucleus:
$$r=r_{0}A^{1/3}$$

Now I'm confused about which is appropriate!

2. Nov 14, 2007

### Dick

The impact parameter is 0. It's heading straight for the nucleus. At closest approach when it's velocity is zero, the potential energy equals the initial kinetic energy. It's not that complicated.

3. Nov 14, 2007

### genloz

So you're saying 15Mev = (3/2) 1.44 Z1Z/r MeV fm
15Mev = (3/2) 1.44 Z1Z/r MeV fm
r=18*Z1Z/125
r=18*12*2/125=3.456fm
?

4. Nov 14, 2007

### Dick

Yes, something like that. Where is the 3/2 coming from?

Last edited: Nov 14, 2007
5. Nov 15, 2007

### genloz

6. Nov 15, 2007

### Dick

That's the energy a particle needs to pass through a charged sphere of radius R. It's not a distance of closest approach. I think you should use the previous formula which treats the particles as points.

7. Nov 16, 2007

### genloz

Okay, thanks... So
U = 1.44 (Z1Z/r) MeV fm
15Mev = 1.44 Z1Z/r MeV fm
r=1.44*Z1Z/125
r=1.44*12*2/125
r=0.276fm

But this is even smaller than the radius of the nucleus which doesn't really make sense...

8. Nov 16, 2007

### Dick

Now where did the 125 come from? Isn't that supposed to be a 15?

9. Nov 16, 2007

### genloz

Oh, you're right.. sorry, so:
r=2.3fm...

That's still smaller than r=r0A^(1/3)=1.3*12^(1/3)=2.9762 (radius of the atom)... so that means it's closer than the electrons to the nucleus?
Also what happened to coulomb repulsion?

10. Nov 16, 2007

### Dick

You just computed the radius of the nucleus, not of the atom. The atom is MUCH bigger. I guess that's telling you that a 15MeV alpha has enough kinetic energy to overcome the repulsion to the degree it can interact with the nucleus. The electrons don't have much to do with it, almost all of the serious repulsion takes place inside of the electron clouds.

11. Nov 16, 2007

### genloz

Ok, thanks very much... so the formula U = 1.44 (Z1Z/r) MeV fm already takes the coulomb force into consideration?