# Rutherford Scattering Problem.

1. Apr 3, 2010

### Cilabitaon

1. The problem statement, all variables and given/known data
In an experiment, an alpha particle of mass 6.7e-27 kg; charge 3.2e-19C travels in a straight line with a speed 2.0e7m/s towards the centre of a gold nucleus.

Find:
1. The kinetic energy($$KE$$) of the alpha particle;
2. The potential(V) at a distance '$$x$$' from the centre of the gold nucleus(of charge 1.3e-17C)
3. The closest distance of approach, and prove that it is 2.8e-14m

2. Relevant equations
$$KE=\frac{mv^{2}}{2}$$
$$E=\frac{F}{Q}$$
$$E=\frac{V}{x}$$

3. The attempt at a solution
1. $$KE=\frac{mv^{2}}{2} = \frac{1}{2}·{(6.7*10^{-27}) · (2*10^{7})^{2}}$$ = $$_1.34*10^{-12}J$$
This part I'm pretty sure of, and I know I should be doing these in my sleep by now, but it was 3:00am when I did this and haven't checked any of these much.

2. E=F/Q and E=V/x

F/Q=V/x -> V=(Fx)/Q

I really had no idea what I was being asked here; yes I realise it tells me clear-as-day, but I just can't seem to get anywhere with this question.

2. I really didn't have a clue. I have tried so many approaches with each being as tame as the last: I first tried initial KE = final PE, to no avail; but that may just be since every time I go over the question I end up answering it in the same WRONG way.

Sorry, I'm not asking anyone to do my homework for me, but I just feel like I'm missing that little piece of info I happened to overlook.

Last edited: Apr 4, 2010
2. Apr 4, 2010

### ehild

For question 1, your result for the KE of the alpha particle is correct.

Question 2. is about the potential of the gold nucleus. It is considered a point charge of 1.3e-17C. What is the potential at a distance x from a point charge of q?

Question 3 is not a question. Are you sure you copied the problem well? Maybe, you are to find the original distance of the particle from the nucleus.

ehild

3. Apr 4, 2010

### Cilabitaon

I still don't get the question, the point charge q has a value, so am I expected to put that into a formula with potential(V) and distance (x): where all I'd get is (Wx/q)=V

Question 3 is trying to get me to use current data to find the closest distance of approach, but the question tells me what it is just so I can say "yeah, I'm right/wrong" when I get to the final answer.

4. Apr 4, 2010

### ehild

You can find the distance of closest approach if you know the initial position of the particle. Maybe it was meant that the particle started from very far, practically infinity. Or it is asked from where the particle had started, so as the closest approach is at the given distance.

Try to assume that the particle has that 1.34e-12 J KE at infinity. At the closest approach, the KE is converted to potential energy. What is the potential energy when the alpha particle is at a distance x from the gold nucleus? You need to use the Coulomb potential.

ehild

5. Apr 4, 2010

### Cilabitaon

So I need $$V=\frac{Qq}{4\pi\epsilon_{0}r^{2}}$$?

But then that would make $$Q$$ my Au nucleus and my point charge the alpha particle, correct?

Also, I change $$r^{2}$$ for $$x^{2}$$, don't I?

Thanks a lot for this help by the way; it's much appreciated.

6. Apr 4, 2010

### ehild

That is the Coulomb force, not the potential energy!
Yes, r or x is the distance between the nucleus and the alpha particle. And Q is the charge of the gold nucleus and q is that of the alpha particle.

ehild

7. Apr 4, 2010

### Cilabitaon

Sorry, I was quoting from my A2 textbook and it states:
Force between two point charges;
$$F=\frac{Q_{1}Q_{2}}{4\pi\epsilon_{0}r^{2}}$$

and

electric potential;
$$V=\frac{W}{Q}$$
$$V=\frac{Qq}{4\pi\epsilon_{0}r^{2}}$$

Should the latter not have $$r$$, not $$r^{2}$$?

I still don't know the potential at any point, or the distance, so how can potential energy changes be used?

8. Apr 4, 2010

### Cilabitaon

I've just thought, with the equation $$V=\frac{W}{Q}$$ I can find $$V$$ for the $$\alpha$$ particle.

$$W=1.34*10^{-12}J$$

$$V=\frac{1.34*10^{-12}}{3.2*10^{-19}} = 4.1875*10^{6}JC^{-1}$$

9. Apr 4, 2010

### ehild

The potential around a point charge Q is

$$V=\frac{Q}{4\pi\epsilon_{0}r}$$

What you cited V=W/Q means the definition of the potential: It is the work done by the electric field on unit positive charge when it moves from the given point to infinite, or to a position where the potential is zero. The potential energy of a charge q is Vq.
So, find the distance from the nucleus, where the potential is

$$V= 4.1875*10^{6}JC^{-1}$$

ehild

10. Apr 4, 2010

### Cilabitaon

Thanks a lot!

This does give me confidence, but now I'm wondering; isn't this the potential of the $$\alpha$$ that we have both just found?

edit:

Oh my goodness, I do believe I am an idiot :)

Thanks so much, I've just realised I've been making the same problems for myself every time I go through this, and now it's just become clear that it wasn't a problem at all!

$$V = \frac{Q}{4 \pi \epsilon_{0}r}$$

$$r = \frac{Q}{V \cdot 4 \pi \epsilon_{0}}$$

$$r = \frac{1.3 \times 10^{-17}}{(4.1875 \times 10^{6}) \times 4 \pi \times (8.85 \times 10^{-12})}$$

$$r = 2.79 \times 10^{-14}m$$

*problem solved* ;)

Last edited: Apr 4, 2010
11. Apr 4, 2010

### ehild

At last :rofl:

ehild