Calculating $\alpha$-Particles Detected During Rutherford Scattering (Cross Sections)

I am currently studying nuclear cross sections and this question considers the classic Rutherford scattering experiment. Here is a diagram of the experiment: The detector is at right-angles to the source and since to keep things simple the solid angle coverage of the detector is 1 steradian.

My task is to calculate the number of $\alpha$-particles that are detected in an hour. I have calculated the differential cross section and was planning to multiply this by the exposure (which I'll change to per hour), the target density and the efficiency of the detector.

My lecturer tells me that I need to also multiply by the flux. My question is: how do I go about calculating the flux in this situation?

Any advice would be much appreciated, Sean.

N.B. Sorry for the wordiness of this post.

• Start by reminding your self what a "flux" is in general. Then think about what flux your instructor meant. – dmckee Nov 9 '14 at 16:23
• If it's flux of $\alpha$-particles hitting/scattering off of the foil and I assume that all of the particles hit the foil and are scattered, would that meant that the flux is simply 1. Ergo in the calculation above about multiplying by it is simply arbitrary? – Vielbein Nov 9 '14 at 16:38
• Reading your comment and the text of the question again, I think that it is possible that you and your instructor have crossed wires on the vocabulary front. Compare the definitions of exposure and flux. – dmckee Nov 9 '14 at 16:50
• Hmm that would make some sense, I've now got an answer in units of grams/hour (grams comes from the density). This doesn't seem right, would you recommend dividing by the mass of an $\alpha$-particle or have I done something wrong? Thanks for the help by the way. – Vielbein Nov 9 '14 at 17:15

Let the number incident/unit time = $N_I$, and the number detected/unit time = $N_D$
We need to consider the number of gold nucleons per unit area in the target. To find this we do: $\rho \over m_{Au}$
The flux of the incident particles, $J$ is simply $J$ = $N_I$
Let the eefficiency of the detector = $\eta$ $$N_D = {d\sigma \over d\Omega}~ \Omega~ {\rho\over m_{Au}}~N_I~\eta$$ So the total detected, $D$, over some interval, $t$, is: $$D = N_D~t={d\sigma \over d\Omega}~ \Omega~ {\rho\over m_{Au}}~N_I~\eta~t$$
• You might want to specify the meaning and units of $\rho$ a little better. Or perhaps use $\sigma$ which is somewhat more conventional for areal densities. – dmckee Nov 14 '14 at 2:36