Rutherford Scattering: Troubleshooting Wrong Atomic Number

[quietrain]

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!

 

[Bill_K]

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.

 

[quietrain]

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?

 

[quietrain]

also, i don't understand how the scattering formula can become this one

from hyperphysics

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 :(

 

[jtbell]

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

$$dN = Nndxd\Omega\frac{d\sigma}{d\Omega}$$

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

$$d\Omega = \frac{dA}{r^2}$$

so

$$dN = Nndx\frac{dA}{r^2}\frac{d\sigma}{d\Omega}$$
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. $d\sigma/d\Omega$ is the proportionality constant. It can vary with the direction the the particles are scattered into (angle $\theta$ from the original beam axis, and angle $\phi$ azimuthally around the beam axis), and with the energy of the incoming particles.

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

 

[quietrain]

ah i see ! thanks!