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

• Taylor_1989
In summary, the conversation discussed an experiment on Rutherford scattering and the necessary corrections needed to compare the results with Rutherford's scattering formula. The correction factor was determined using a figure and calculations were done to find the solid angle of the ring. The conversation also touched on the measurement of the number of incident particles per unit area and how it could further quantify the results.
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

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).

Last edited:
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 alot, for me.

## What is Rutherford scattering?

Rutherford scattering is a phenomenon in which alpha particles, a type of positively charged particle, are deflected when they pass through a thin foil of gold or another heavy element. This experiment was conducted by Ernest Rutherford in 1911 as part of his research on the structure of atoms.

## What is the significance of Rutherford scattering?

The results of Rutherford's experiment were unexpected and revolutionized our understanding of the structure of atoms. It showed that atoms have a small, dense, positively charged nucleus at their center, with most of the atom being empty space. This was a major departure from the previously accepted "plum-pudding" model of the atom.

## How does Rutherford scattering work?

In Rutherford scattering, alpha particles are fired at a thin gold foil. The majority of the particles pass through the foil with little to no deflection, but a small percentage are deflected at large angles. This is because the alpha particles encounter the positive charge of the nucleus, causing them to change direction.

## What factors affect the results of Rutherford scattering?

The results of Rutherford scattering can be affected by several factors, including the charge and size of the nucleus, the distance between the alpha particles and the nucleus, and the energy of the alpha particles. Additionally, the thickness and composition of the foil can also impact the results.

## What is the modern interpretation of Rutherford scattering?

The modern interpretation of Rutherford scattering is based on the understanding of atomic structure and the forces between charged particles. The positively charged nucleus exerts a repulsive force on the positively charged alpha particles, causing them to scatter at large angles. This phenomenon is now used in various fields, including nuclear physics and medical imaging.

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