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!