Register to reply

Python and Rutherford Scattering

by maximus123
Tags: python, rutherford
Share this thread:
maximus123
#1
Dec10-12, 04:40 AM
P: 29
Hello everyone

I have been supplied with this eqn

[tex]n=\frac{Nat}{16r^2}(\frac{2Ze^2}{4\pi\epsilon_0E_K})^2cosec^4(\frac{\ph i}{2})[/tex]

for Rutherford scattering.

[tex]N[/tex] is the number of alpha particles incident on a unit area of foil

[tex]t[/tex] is the thickness of the foil

[tex]a[/tex] is the number of atoms per unit volume within the foil

[tex]r[/tex] is the distance of the detector away from the collision

[tex]Z[/tex] is obviously the atomic number of the foil's material

and [tex]E_k[/tex] is the kinetic energy of the incident alpha particles

I need to use python to plot a graph of number of counts 'n' versus the scattering angle phi. To do this I need to create a GUI with entry fields for some of the above variables. This bit is not the problem.
The problem is so far I cannot get the eqn to give a useful plot.These figures are not based on an actual experiment which I have had to carry out. These are all theoretical values for the purpose of the computer program.
Could anyone suggest reasonable values for the above variables. Z is not necessary as I'm using 79 (gold). For example I've no idea how many atoms per unit volume there would be in a sheet of foil prepared for one of these experiments. Then if these reasonable values don't plot well either I will know that it is my code that needs looking at (which I'm sure is fine at the mo).
Any help would be greatly...greatly appreciated.
Phys.Org News Partner Science news on Phys.org
'Smart material' chin strap harvests energy from chewing
King Richard III died painfully on battlefield
Capturing ancient Maya sites from both a rat's and a 'bat's eye view'
uart
#2
Dec10-12, 06:43 AM
Sci Advisor
P: 2,751
Quote Quote by maximus123 View Post
For example I've no idea how many atoms per unit volume there would be in a sheet of foil prepared for one of these experiments.
You can work that out easily enough from the density and the atomic mass. The atomic mass is 197 grams per mole, and the density is approx 19.3 grams per cm^3.

So there are 6.02E23/197 atoms per gram, and therefore 6.02E23/197*19.3 atoms per cm^3. This comes to about 5.9E22 by my quick calculation.
uart
#3
Dec10-12, 06:57 AM
Sci Advisor
P: 2,751
Quote Quote by maximus123 View Post
The problem is so far I cannot get the eqn to give a useful plot.These figures are not based on an actual experiment which I have had to carry out. These are all theoretical values for the purpose of the computer program.
Well all of those constants essentially just boil down to a scaling factor for the cosec(theta/2) term anyway. So they are in effect irrelevant to the actual shape of the plot.

My problem with that equation is that the density of strikes goes to infinity as theta goes to zero, which doesn't seem right. Unless it is an approximation that is only valid over a limited range of theta perhaps. Do you have any more information on the origins of that formula?

gsal
#4
Dec10-12, 09:50 PM
P: 898
Python and Rutherford Scattering

like uart said, the basic shape of the plot is not going to change if all you are doing is plotting phi vs n.

here is what it looks like using Wolfram.
maximus123
#5
Dec12-12, 08:14 AM
P: 29
Thanks a lot uart. Not sure how my department has come up with the equation, tis probably some approximation for the purposes of the python exercise we have to do. You're right about the division by zero, I've had to make sure my range of angles does not include zero (which is stupid because most angles in this experiment would be zero)

Any way I achieved my plot, an exponentially decaying curve.

Thanks again.
jtbell
#6
Dec12-12, 08:24 AM
Mentor
jtbell's Avatar
P: 11,864
Quote Quote by maximus123 View Post
Thanks a lot uart. Not sure how my department has come up with the equation, tis probably some approximation for the purposes of the python exercise we have to do.
It looks like a standard Rutherford scattering formula that you can find in textbooks and on Wikipedia. The result n is probably the number of scattered particles per m2 of detector area. It assumes point-like projectiles (alpha particles) and targets (nuclei).


Register to reply

Related Discussions
Rutherford Scattering -scattering angle and formula Introductory Physics Homework 0
Rutherford scattering Advanced Physics Homework 0
Rutherford Scattering Introductory Physics Homework 1
Rutherford Scattering Advanced Physics Homework 1
Rutherford Scattering Introductory Physics Homework 1