Calculating Scattering Cross Sections

In summary, when a particle beam of alpha particles of kinetic energy 5.3 MeV and intensity 10^4 particle/sec hits a gold foil with a thickness of 1x10^-5 cm, a particle counter of area 1 cm^{2} is placed on the opposite side of the foil from the incoming beam at a distance of 10 cm. If the angle between the center of the foil and the center of the detector is 10 degrees and 45 degrees, the counter should register 11.4 and 4020 counts per second, respectively.
  • #1
strangequark
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0

Homework Statement



A beam of [tex]\alpha[/tex]-particles, of kinetic energy 5.3 MeV and intensity [tex]10^{4}[/tex] particle/sec, is incident normally on gold foil with thinckness [tex]1 x 10^{-5} cm[/tex]. (The density, atomic weight and atomic number of gold are 19.3 g/cm, 197 and 79 respectively.) A particle counter of area [tex]1 cm^{2}[/tex] is placed on the opposite side of the foil from the incoming beam at a distance of 10 cm. Suppose [tex]\theta[/tex] is the angle between the center of the foil and the center of the detector. How many counts should we expect per second if the detector registers all particles passing through it at [tex]\theta=10 degrees[/tex] and [tex]\theta=45 degrees[/tex]?

Homework Equations



[tex]\frac{d\sigma}{d\Omega}=(\frac{1}{4\pi\epsilon_{0}})^{2} (\frac{zZe^{2}}{2Mv^{2}})^{2} I n \frac{1}{sin^{4}(\frac{\theta}{2})} [/tex]

where n is the number of nuclei per unit area, and I is the incident intensity.

also told that:

[tex]2 \pi sin(\theta) d\theta=d\Omega [/tex]

and

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

The Attempt at a Solution



First I calculated the constant:

[tex](\frac{1}{4\pi\epsilon_{0}})^{2} (\frac{zZe^{2}}{2Mv^{2}})^{2}[/tex]

where I had [tex]2Mv^{2}=4(5.3 MeV) [/tex], [tex]z=2, Z=79[/tex]

Then I found n:

[tex]n=\frac{19.3g}{cm^{3}}\frac{1}{197amu}1 x 10^{-5}cm [/tex]

now's where I have a problem... I'm trying to integrate:

[tex]2\pi(\frac{sin(\theta)}{sin^{4}(\frac{\theta}{2})})d\theta[/tex]

but I don't know what angles to integrate thru... when I draw a diagram, I get a pretty small angular range and my answer comes out too small.


the answers are:

for [tex]45 degrees = 11.4 counts/sec. [/tex]
and for [tex] 10 degrees = 4020 counts/sec. [/tex]

I could really use some help on seeing how these answers are obtained... thanks!
 
Last edited:
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  • #2
Haven't looked in detail at your attempt. First, make sure you're using the correct units, and consistently. Usually safest to work in SI units unless an equation specifies to use some non-standard units. Next, your integration region is indeed the small angular width of the detector (10deg-small amount to 10deg+small amount and similarly for 45deg case). For the integration, you can define [itex]\theta'=\theta/2 [/itex] and use the trig identity [itex]\sin(2\theta')=2\sin\theta' \cos\theta'[/itex]. Then [itex]d\theta=2d\theta'[/itex] and the cosine can be "brought into the differential": [itex]
\cos\theta' d\theta'=d \sin\theta'[/itex].
 
  • #3
i am doing that exact integral now. i just posted it in cal.analysis. i am trying to use the substitution u=cos(theta/2) but it hasnt produced anything of value yet.
 
  • #4
ok i finally finished this problem. you have to find d(omega), not by integrating the sin stuff u have above. use d(omega) = INTEGRAL of dA/r^2 so you pull out the r^s terms using 10^2 cm^2, and use 1 cm for the area A. youll get 1/100.second thing is you have to correct your n value. i converted the density to kg/m^2. then i divided it 197, and multiplied it by avagad's number to arrive at 5.897x10^28. crunching all this then youll get close. you should get 6...x10^-5 x 1/sin^4(theta/2). the last trick ill leave for you to do. remember you have to find the counts per hr. but the equation will be in s^-1.
good luck.
 

1. What is a scattering cross section?

A scattering cross section is a measure of how likely it is for a particle or wave to be scattered when it interacts with a target. It is typically expressed in units of area and is used to describe the probability of a scattering event occurring.

2. How is a scattering cross section calculated?

The scattering cross section is calculated by taking the ratio of the number of scattered particles to the number of incident particles and multiplying it by the area of the target. Other factors such as the energy and angle of the incident particles may also affect the calculation.

3. What is the significance of scattering cross sections in physics?

Scattering cross sections are important in understanding the properties and behavior of particles and waves. They can provide information about the size, shape, and composition of targets, as well as the strength and nature of interactions between particles.

4. Are scattering cross sections constant or do they vary?

Scattering cross sections can vary depending on the type of interaction, energy of the particles, and characteristics of the target. They may also change with different experimental conditions, such as temperature or pressure.

5. How are scattering cross sections used in research and applications?

Scattering cross sections are used in a variety of research fields, including particle physics, astrophysics, and materials science. They are also utilized in practical applications such as medical imaging and environmental monitoring.

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