Impact Parameter, Closest Approach

[genloz]

Homework Statement

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?

Homework 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}}

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!

 

[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.

 

[genloz]

Okay.. that's helpful, thanks...

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
?

 

[Dick]

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

 

[genloz]

Well I found the formula for kinetic energy:
Umax = (3/2) 1.44 Z1Z/R MeV fm
on this website:
http://www.upscale.utoronto.ca/GeneralInterest/DBailey/SubAtomic/Lectures/LectF07/Lect07.htm

 

[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.

 

[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...

 

[Dick]

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

 

[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?

 

[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.

 

[genloz]

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