Rutherford scattering calculation

[strangequark]

Homework Statement

The fraction of [tex]6.0 MeV[/tex] protons scattered by thin gold foil, of density [tex]\rho=19.3 g/cm^{3}[/tex], from the incident beam into a region where scattering angles exceed [tex]60 degrees[/tex] is equal to [tex] 2.0 x 10^{-5} [/tex]. Calculate the thickness of the gold foil using the result of the previous problem.

Homework Equations

The result I got for the previous problem was a formula for the number of particles scattered thru and angle of [tex]\theta[/tex] or greater and is:

[tex]N(\theta)= (\frac{1}{4\pi\epsilon_{0}})^{2} \pi I \rho t (\frac{zZe^{2}}{Mv^{2}})^{2} cot^{2}(\frac{\theta}{2}) [/tex]

The Attempt at a Solution

So I took [tex]\frac{N(\theta)}{I}[/tex] which is equivalent to removing the I from the above equation and attempted to solve for t, which is the thickness...

[tex]t=\frac{2x10^{-5}}{(\frac{1}{4\pi\epsilon_{0}})^{2} \pi (19.3 g/cm^{3}) (\frac{79(1.602x10^{-19}coul)^{2}}{2(6 MeV)})^{2} cot^{2}(\frac{\pi}{6})}[/tex]

problem is that I keep getting an answer in m/kg, so I must be missing some term, though I can't figure out what it is...

any help is very much appreciated.

 

[Astronuc]

Absorber thickness is sometime measured in kg/m² or gm/cm² such that dividing by density in kg/m³ or gm/cm³ yields a thickness in m or cm, respectively.

Make sure the relationship between N and I is correct, and check one's units on the various terms.

 

[ogb p]

Your attempt at a solution is on the right track, but there are a few things that need to be corrected in order to get the correct units for thickness. First, the units for the electric field constant, \epsilon_{0}, should be factored in to the equation. It has units of \frac{C^{2}}{N\cdot m^{2}}. Additionally, the units for the mass, M, should be in kg, and the units for the velocity, v, should be in m/s. Finally, the units for the scattering angle, \theta, should be in radians.

So the corrected equation for thickness would be:

t=\frac{2x10^{-5}}{(\frac{1}{4\pi\epsilon_{0}})^{2} \pi (19.3 g/cm^{3}) (\frac{79(1.602x10^{-19}coul)^{2}}{2(6 MeV)})^{2} cot^{2}(\frac{\pi}{6}) \frac{m^{3}\cdot kg^{2}}{C^{4}}}

Using the conversion factors, 1 J = 1 kg m^{2}/s^{2} and 1 MeV = 1.602x10^{-13} J, the final answer for the thickness would be in meters.

I hope this helps clarify the issue and leads you to the correct solution. Good luck with your calculations!