# Rutherford Scattering

1. Dec 29, 2015

### says

1. The problem statement, all variables and given/known data
Calculate the cross section for the scattering of a 10 MeV alpha particle by a gold nucleus (Z=79, A=197) through an angle greater than a) 10 degrees b) 20 degrees c) 30 degrees. Neglect Nuclear recoil.

2. Relevant equations
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/crosec.html

3. The attempt at a solution
I've seen a number of equations that relate to rutherford scattering. The one on hyperphysics looks neat, but it doesn't say what the variables in the equation are. I don't know what the small 'ke' means and what the two angles are supposed to be in the equation either. I'm assuming they are both the scattering angle.

2. Dec 29, 2015

### theodoros.mihos

You can find that by dimension check.

3. Dec 29, 2015

### says

small ke is Newton / metre? I assume it has something to do with the Coulomb Force / Impact Parameter.

4. Dec 29, 2015

### theodoros.mihos

$$\frac{k\,e^2}{KE}$$
must be on meters.

5. Dec 29, 2015

### says

kg/s2 on the numerator would give units of barns

6. Dec 29, 2015

### says

ahhhh! I was thinking in the wrong units. So I found that ke2 = 1.44 MeV*fm

If I plug that into the cross section equation given by hyperphysics I get an answer of 0.022813 fm2

But my textbook says the answer is 5.3*104 fm2

7. Dec 29, 2015

### Staff: Mentor

Try converting the given kinetic energy (KE) into joules. Then you can use standard SI (MKS) units for the rest. k is the Coulomb constant from Coulomb's Law: k = 1/(4πε0). e is the elementary charge (magnitude of charge on electron or proton).

Last edited: Dec 29, 2015
8. Dec 29, 2015

### says

I still get the incorrect answer.

k= 8.98×109
e = 1.60x10-19

I use those values and plug it in to the equation and I don't get 5.3*104 fm2

9. Dec 29, 2015

### Staff: Mentor

What did you use for the KE? (in J)

10. Dec 29, 2015

### says

1.60218*10-12

11. Dec 29, 2015

### Staff: Mentor

Your input numbers agree with mine, but I do get the textbook's answer (for 10°). Check your arithmetic again, and if you still can't find the error, show exactly how you set it up, with the numbers substituted into the equation.

12. Dec 29, 2015

### says

π * 22 * ( 8.9*109 * ((1.6*10-19)2) / 1.6*10-12 ) (1+cos(10) / 1-cos(10) )

= 3.3*10-29

Sorry about the formatting, but I've got it in degree mode and I've checked all the brackets etc.

13. Dec 29, 2015

### says

And my units seem right with dim. analysis

N*m2/C2 * C2 / kg * m2*s-2

14. Dec 29, 2015

### Staff: Mentor

Z is for gold (79), not the alpha particle (2).

Also, you omitted one power of 2. Look at the original equation carefully.

Finally, you may have some discrepancy from roundoff error because you used only two significant figures.

15. Dec 29, 2015

### says

oh, damn! I should've seen that. I tend to look at equations and hope that someone has wrote, where Z= ...., k=...., because sometimes I either forget things, i.e. like coulomb constant and elementary charge, and I find it easier to understand relationships with equations instead of a slab of text. Thanks for your help! :)

16. Dec 29, 2015

### says

adding the power of 2 and changing the Z to 79, i get 5.19e-26

17. Dec 29, 2015

### says

I'm still out by some order of magnitude, but I'm not sure why.

18. Dec 29, 2015

### says

my answer is in m, and not m2. If I square my answer though it's out by an even greater magnitude...

textbook = 5.3*104 fm2

me = 5.19 *10-26 m = 5.19*10-11 fm

19. Dec 29, 2015

### says

Figured it out. I replaced the units with MeV and got the correct answer.