# Rutherford Scattering Problem.

• Cilabitaon
In summary, in an experiment, an alpha particle of mass 6.7e-27 kg and charge 3.2e-19C travels in a straight line with a speed of 2.0e7m/s towards the center of a gold nucleus of charge 1.3e-17C. The kinetic energy of the alpha particle is 1.34e-12J. The potential at a distance x from the center of the gold nucleus is V=4.1875*10^6JC^-1. The closest distance of approach is 2.79e-14m.

## Homework Statement

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

## Homework Equations

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

## 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 realize 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.

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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

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.

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

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.

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

ehild said:
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

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?

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

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

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

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

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 realized 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* ;)

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At last :rofl:

ehild