How Can I Calculate the Differential Cross-Section for Bremsstrahlung Emission?

[Jock Flannigan]

Hello guys!

I was wondering if you can help me out with the differential cross section. We set up an experiment to measure the DCS in gold of the bremsstrahlung emission by electrons. The goal is to determine:

$\frac{d\sigma}{dW d\Omega}$ (1)​

for a 1.9 MeV electron beam at an angle of emission of 30$^{\circ}$. However only for when the energy of the emitted photon is around 720 keV.
I know the beam electric current is constant during the duration and equals 0.3 $\mu A$, the target is a gold slab of mass thickness 130 $\mu g /cm^{2}$ with a mass density $19.32 g / cm^{3}$. The detector is 1.3m from the target. The collimator aperture is circular with a radius of 0.240 cm. The beam is on for 5min. The number of photons detected is 1423.

I am not asking to solve it, but if you can give me few hints on how to do it. Also if you can explain the individual elements of the equation (1)

Thank you very much!

 

[mfb]

Did your detector take care about the energy range? What is the range of energies it covered?

Every book about particle physics covers those basic differential cross-section problems in an early chapter. The equation does not really have individual elements - it is the cross-section per energy range (of the photon I guess) and solid angle.
As it is just a large multiplication/division of all the quantities, the question is very homework-like. I moved this thread to the homework question.