Calc Nuclear Cross Section for Alpha Particles Reacting with 48-Ca

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Matt atkinson
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Alpha particles can react with 48-Ca to produce protons. Consider a beam of alpha particles of current 20nA fired at a think target of Ca of thickness $$1mg/cm^2$$. A detector subtending a solid angle of 4*10^-3 steradians, records 20 protons per second. Determine the total cross section (in mb) for the reaction. State any assumptions made.

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So if I=20nA then the initial rate of alphas is:
$$R_o=\frac{I}{2e}=6.25 \times 10^{10} atoms / sec$$
And also the number of atoms in the think foil per $m^2$ is $$N=1.25\times 10^{23} atoms/m^2$$.
But I have no idea what to do from here I understand I'm given rate of initial alphas and rate of detection and a solid angle.
Could someone just point me in the right direction?
 
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Assuming emission to be a semi-sphere and which is 2pi steradians then the protons per second created would be 31415 per second created so that means that there were that many reactions.
The only equation I can find is
$$\sigma = \frac{R}{I} $$
Reactions per second, over incident particles per second per unit area.
I know I calculated the number of alpha Per second but I don't quite understand how to get it per unit area and what the the number of 48-Ca atoms has to do with anything. I'm probably missing something simple.
 
Oh if i instead workout the number of particles per uni cm^2 and then take the beam to have a area of 1cm^2 ?
 
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I'm really quite stuck, hmm this is what I've done I just don't know how to put it together.

Let me know if the image can't be read, I will type my workings in latex.
ImageUploadedByPhysics Forums1426600456.547267.jpg
 
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