Rutherford Scattering: Troubleshooting Wrong Atomic Number

In summary, the Rutherford scattering experiment measures the fraction of particles scattered into a particular solid angle relative to the incoming beam, known as the differential cross section. The left hand side of the equation does not equal the count rate N, as it is missing a factor of beam intensity I. This intensity can be calculated through the source of radioactive decay. The formula also takes into account the number of particles, the number density of target particles, the thickness of the target, and the distance between the target and detector. For larger detectors, the formula must be integrated over the angles subtended by the detector.
  • #1
quietrain
655
2
according to wiki's rutherford scattering experiment,

"The fraction of particles that is scattered into a particular solid angle at a given direction relative to the incoming beam is called the differential cross section and is given by
[URL]http://www.advancedlab.org/mediawiki/images/math/5/d/4/5d494974ec5febddca8376c87d0338ae.png"[/URL]

does the left hand side = count rate N?

so if i were to plot lg N vs lg sin(θ/2)
so i get lg N = -4lg sin(θ/2) + lg Constant

so my Constant = those (ZZe2/4E)2 ?

but when i calculate atomic number of gold from my constant, it is not 79, it's some insanely big number :(

e = electron charge , E = kinetic energy of alpha particles i used 5MeV, are these right? my source is radioactive decay of 241-Americium

does anyone know where i went wrong?
thanks!
 
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  • #2
does the left hand side = count rate N?
No, you're missing a factor. You can pretty much tell what's missing by checking the dimensions. What's the dimensions of the right hand side? Remember ZZe2/r is the Coulomb potential energy, so ZZe2/E must have the dimensions of a length. Ergo the RHS has dimensions length2, or area. Aha, maybe that's why they call it a cross-section!

What you're looking for is the count rate, N = no of particles/sec. To get something having these dimensions you need an extra factor "no of particles/sec/area" This quantity is I, the beam intensity. Obviously if you double the intensity of your source you'll double the count rate. Your formula should read N = I dσ/dΩ.
 
  • #3
Bill_K said:
No, you're missing a factor. You can pretty much tell what's missing by checking the dimensions. What's the dimensions of the right hand side? Remember ZZe2/r is the Coulomb potential energy, so ZZe2/E must have the dimensions of a length. Ergo the RHS has dimensions length2, or area. Aha, maybe that's why they call it a cross-section!

What you're looking for is the count rate, N = no of particles/sec. To get something having these dimensions you need an extra factor "no of particles/sec/area" This quantity is I, the beam intensity. Obviously if you double the intensity of your source you'll double the count rate. Your formula should read N = I dσ/dΩ.

Ah i see!

but how do i calculate the value of beam intensity I? if my experiment only gives me a source of 241-Americium radioactive decaying and producing alpha particles?
 
  • #4
also, i don't understand how the scattering formula can become this one

from hyperphysics
ruteq2.gif


its as though telling me that

NnLk2/4r2 = Z2

but if we add in the intensity like you suggested earlier, then it becomes

NnLk2/4r2 = Z2 N/L2 , assuming L2 is area

so only if this equation above is true, then we get the equation in the first post. but how is this equation as such bewilders me :(
 
  • #5
quietrain said:
according to wiki's rutherford scattering experiment,

"The fraction of particles that is scattered into a particular solid angle at a given direction relative to the incoming beam is called the differential cross section and is given by
[URL]http://www.advancedlab.org/mediawiki/images/math/5/d/4/5d494974ec5febddca8376c87d0338ae.png"[/URL]

does the left hand side = count rate N?

The operational definition of [itex]d\sigma/d\Omega[/itex] is as follows: If you have N projectile particles (alphas in this case) coming into a "thin" target of thickness dx which contains [itex]n[/itex] targets (nuclei in this case) per m^3, and dN of the scattered projectiles emerge into a "small" solid angle [itex]d\Omega[/itex], then

[tex]dN = Nndxd\Omega\frac{d\sigma}{d\Omega}[/tex]

[itex]d\Omega[/itex] is determined by the cross-sectional area dA of your detector and its distance r from the target:

[tex]d\Omega = \frac{dA}{r^2}[/tex]

so

[tex]dN = Nndx\frac{dA}{r^2}\frac{d\sigma}{d\Omega}[/tex]
Verbally: the number of particles entering your detector is proportional to the the number of incoming particles, the number density of target particles, the thickness of the target, the cross-section area of your detector, and inversely proportional to the square of the distance between the target and the detector. [itex]d\sigma/d\Omega[/itex] is the proportionality constant. It can vary with the direction the the particles are scattered into (angle [itex]\theta[/itex] from the original beam axis, and angle [itex]\phi[/itex] azimuthally around the beam axis), and with the energy of the incoming particles.

This is for a "small" detector such that [itex]d\sigma/d\Omega[/itex] is effectively uniform over its area. For "large" detectors you have to take into account the variation with angle, by integrating over [itex]\theta[/itex] and [itex]\phi[/itex] subtended by the detector.

Also this is for targets that are "thin" along the incoming beam direction. For "thick" targets N decreases as the beam proceeds through the target, and you have to take that into account by integrating over x.
 
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  • #6
ah i see ! thanks!
 

FAQ: Rutherford Scattering: Troubleshooting Wrong Atomic Number

1. What is Rutherford scattering and its significance in atomic structure?

Rutherford scattering is a phenomenon in which alpha particles, small positively charged particles, are scattered as they pass through a thin foil. This experiment led to the discovery of the atomic nucleus and provided evidence for the existence of a small, dense, positively charged nucleus within the atom.

2. How is Rutherford scattering used to determine the atomic number of an element?

Rutherford scattering can be used to determine the atomic number of an element by measuring the angle at which the alpha particles are scattered. The higher the atomic number, the greater the repulsion between the alpha particles and the nucleus, resulting in a larger scattering angle.

3. What are some common mistakes that can lead to wrong atomic number results in Rutherford scattering?

Some common mistakes in Rutherford scattering that can lead to wrong atomic number results include using an incorrect foil thickness, not accounting for background scattering, and failing to properly calibrate the detector.

4. How can these mistakes be troubleshooted to obtain accurate results?

To troubleshoot wrong atomic number results in Rutherford scattering, it is important to double-check the thickness of the foil being used and to account for any background scattering by subtracting it from the total scattering data. Additionally, calibrating the detector properly and repeating the experiment multiple times can help ensure accurate results.

5. Are there any other factors to consider when conducting Rutherford scattering experiments?

Yes, some other factors to consider when conducting Rutherford scattering experiments include the energy and trajectory of the alpha particles, as well as the atomic structure and composition of the thin foil being used. Additionally, controlling for external factors such as temperature and air currents can also impact the accuracy of the results.

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