# Cant seem to figer it out

1. Aug 20, 2006

### fatboy12341

in a rutherford gold experiment, alpha particles with sharge +2e and a kinetic energy of 7.7 Mev were beamed at gold foil. the nuclius of a gold atom contains 79 protons, giving it a charge of +79e. What is the closest distance that an alpha particle can get to a gold nucleus when it approaches head on ?

ok i used the equation En=k^2 x e^4 me . Z^2
2xh^2 h^2

i am getting n^2= -9.07x10^21

i know this is wrong and am getting quite frustrated
if some on could help me it would be great

2. Aug 21, 2006

### quasar987

You asked this question already and got an answer. If you don't understand what Andrew Mason said, you can ask him for further explanations. I for one, don't see how to find the solution without knowing the distance btw the gold sheet and the source of the alpha particles. :(

3. Aug 22, 2006

### Reshma

You know that the distance of closest approach is of the order of the radius of the Gold atom. The formula for the radius is (if you are aware of the derivation):
$$R = \frac{1}{4\pi \epsilon_0}\frac{2Ze^2}{mv^2}$$

If you can see, the KE of the atom has been incorporated in this equation. You just have to substitute the required quantities to obtain the distance of closest approach. You can take it from here I hope...

4. Aug 28, 2006

i don't know but your possibly usingthe wrong units. (ie Joules instead of electron-volts) the solution is straight forward.

kinetic energy = electrical potential

Ke = that's given

Elec Pot = KQq/r

rearrange for r and make sure that k has the correct units.

5. Aug 28, 2006

### quasar987

I don't see how the problem has a solution if we are not given the distance 'd' between the gold sheet and the alpha particle source because, if we take a coordinate system where the origin is on the gold sheet (x-axis perpenducular to it) the potential energy of the "alpha particle-fixed gold nucleus" system is

$$U(x)=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_{\alpha}}$$

The energy is a constant of the motion, hence for any two positions $x_1$ and $x_2$ of the alpha particle, we have

$$U(x_1)+K(x_1)=U(x_2)+K(x_2)[/itex] Taking $x_1=d$, the position just as the alpha particle leaves the source (where $K=K_0=7.7$ MeV) and $x_2=x_m$, the minimum distance between the alpha particle and the gold nucleus (where K=0), we have [tex]\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{d}+K_0=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_m}$$

Solving for x_m requires the knowledge of the distance 'd' btw the source and the gold sheet.

Last edited: Aug 28, 2006
6. Aug 28, 2006

### nrqed

I thought that in that kind of problem, it was implicitly assumed that the initial distance can be approximated to be infinite. In practice, if the energy is given at, say, a couple of meters from the gold nucleus, the initial potential energy will be completely negligible compared to th efinal potential energy (since the distance of approach is *much* smaller than a few meters).