Calculating the Cross Section for Rutherford Scattering

[says]

Homework Statement

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.

Homework Equations

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

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.

 

[theodoros.mihos]

You can find that by dimension check.

 

[says]

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

 

[theodoros.mihos]

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

 

[says]

kg/s² on the numerator would give units of barns

 

[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

 

[jtbell]

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).

 

[says]

I still get the incorrect answer.

k= 8.98×10⁹
e = 1.60x10^-19

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

 

[jtbell]

k= 8.98×109
e = 1.60x10-19

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

 

[says]

1.60218*10^-12

 

[jtbell]

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.

Make sure your calculator is set to use degrees instead of radians for trig functions.

 

[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.

 

[says]

And my units seem right with dim. analysis

N*m²/C² * C² / kg * m²*s^-2

 

[jtbell]

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.

 

[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! :)

 

[says]

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

 

[says]

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

 

[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

 

[says]

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