What is the necessary correction factor for Rutherford scattering in 3D?

[Taylor_1989]

Homework Statement

So I recently carried out an experiment for Rutherford scattering. From my lab script is states the following

'Recording the scattering rate as function of angle

In order to compare your results with “Rutherford’s scattering formula”, you must first make a
correction for the fact that we have only sampled the scattering in the horizontal plane, whereas the
α-particles are actually scattered in a 3D cone. With the aid of Fig. 6, determine the necessary
correction factor. Apply this correction and compare your results (by graphical representation) with
the theoretical curve of the form:

$$f(\theta)=\frac{A}{sin [ ( θ − B ) / 2 ]}[1]$$'

The figure 6

[![Image is of what was given to help derive our correction][1]][1] [1]: https://i.stack.imgur.com/wvPZ5.png

So I calculated the solid angle in the following way:

I took the surface element on a hemisphere which gave the differential are as follows:

$$dA=r^2 sin\theta d\theta d\phi[2]$$

Now I assume this is the area of the detector.

Now to calculate the are of the ring, I realized that $\phi$ would have to change from $0 \rightarrow 2\pi$

Thus by taking the intergral I was able to find the solid angle of the ring

$$\int_0^{2\pi} r^2 sin\theta d\theta d\phi=2\pi r^2 sin\theta d\theta[3]$$

So by takeing the differential solid angle $d\Omega=\frac{da_{ring}}{r^2}$ I was able to calculate the differential solid angle of the ring to be:

$$d\Omega=2 \pi sin\theta d\theta [4] $$

And now this is where I am confused because according to my lecture I am correct in what I have done and the correction is as follows:

$$N_s=2\pi sin \theta \times N_d [5]$$

where $N_s=scatter$ on the solid angle and $N_d$ is the results that I have recored for a 2d plane.

My question is why do this not include the $d \theta$ I can't see why the $d \theta$ neglected. Also I can seem to make the connection to what $d\theta$ is w/r to the experiemtnt?edit: I now think that my $$d\theta=h/r$$ where h is the height of the detector

 

[Charles Link]

The scattering cross section $\frac{d \sigma}{d \Omega}=\frac{(\frac{dN}{d \Omega})}{(\frac{dN}{d \sigma})}$. It appears that your data doesn't contain a measurement of the number of incident particles per unit area $\frac{dN}{d \sigma}$, so that you are just measuring a normalized $\frac{d \sigma}{d \Omega}$ which is $\frac{dN}{d \Omega}=Count/(A_D/r^2)$, (where $Count=Count(\theta)$). Meanwhile $f(\theta)\, d \theta=\int\limits_{0}^{2 \pi}[ \frac{d \sigma}{d \Omega} \sin{\theta} \, d \theta] \, d \phi$ so that $f(\theta)=2 \pi \sin{\theta} \, (\frac{d \sigma}{d \Omega})$ which is proportional to $(2 \pi \sin{\theta}) \, Count(\theta)$. $\\$ (Since it appears to simply be a normalized function, the $2 \pi$ factor appears to be incidental). $\\$ A measurement of the incident beam intensity per unit area$\frac{dN}{d \sigma}$ would allow for further quantification of the results. e.g. if you know the mass of the gold foil, you can calculate the number of gold atoms, with the result that you could estimate the dimensions (cross sectional area $\sigma$ of the individual gold atom scatterer).

 

[Taylor_1989]

The scattering cross section $\frac{d \sigma}{d \Omega}=\frac{(\frac{dN}{d \Omega})}{(\frac{dN}{d \sigma})}$. It appears that your data doesn't contain a measurement of the number of incident particles per unit area $\frac{dN}{d \sigma}$, so that you are just measuring a normalized $\frac{d \sigma}{d \Omega}$ which is $\frac{dN}{d \Omega}=Count/(A_D/r^2)$, (where $Count=Count(\theta)$). Meanwhile $f(\theta)\, d \theta=\int\limits_{0}^{2 \pi}[ \frac{d \sigma}{d \Omega} \sin{\theta} \, d \theta] \, d \phi$ so that $f(\theta)=2 \pi \sin{\theta} \, (\frac{d \sigma}{d \Omega})$ which is proportional to $(2 \pi \sin{\theta}) \, Count(\theta)$. $\\$ (Since it appears to simply be a normalized function, the $2 \pi$ factor appears to be incidental). $\\$ A measurement of the incident beam intensity per unit area$\frac{dN}{d \sigma}$ would allow for further quantification of the results. e.g. if you know the mass of the gold foil, you can calculate the number of gold atoms, with the result that you could estimate the dimensions (cross sectional area $\sigma$ of the individual gold atom scatterer).

Thank you very much, that has cleared up a lot, for me.