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(Created page with "===Moller Differential Cross Section=== Using the equation from [1] <center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{*2}}\left \{\frac{1+cos^4(\frac{\theta^*}{2})}{sin…")
 
 
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We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>
 
We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>
 +
 +
===Different p<sub>2</sub><sup>1</sup> Values===
 +
 +
Using the conversion of
 +
 +
 +
<center><math>\frac{1}{1GeV^2}=.3894 mb</math></center>
 +
 +
 +
<center><math>\sigma=\int d\sigma=\int \frac{d\sigma}{d\Omega_2'}d\Omega</math></center>
 +
 +
 +
The range of the detector is considered to be <math> .10 \le \theta \le .87</math>,<math>-\pi \le \phi \le \pi</math>
 +
 +
 +
<center><math>\sigma=\int_{ .611}^{2.531} \int_{-\pi}^{\pi} \frac{d\sigma}{d\Omega_2'}sin\theta \,d\theta \, d\phi </math></center>
 +
 +
 +
 +
<center><math>\sigma=2\pi \int_{.611}^{2.531} \frac{d\sigma}{d\Omega_2'} sin\theta \,d\theta  </math></center>
 +
 +
 +
 +
<center><math>\sigma=2\pi (1.638)\frac{d\sigma}{d\Omega_2'} </math></center>
 +
 +
 +
 +
<center><math>\sigma=(10.294) \frac{d\sigma}{d\Omega_2'} </math></center>
 +
 +
 +
{| class="wikitable" align="center" border=1
 +
  |+ '''Differential Cross Section Scale for Different p<sub>2</sub><sup>1</sup> Values'''
 +
|-
 +
  ! <math>p_{2}'(MeV)</math>
 +
  ! <math>\frac{d\sigma}{d\Omega_{2}^'}(eV^{-2})</math>
 +
  ! <math>\frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2})</math>
 +
  ! <math>\frac{d\sigma}{d\Omega_{2}^'}(mb)</math>
 +
  ! <math>\frac{d\sigma}{d\Omega_{2}^'}(b)</math>
 +
  ! <math>\sigma(b)</math>
 +
|-
 +
  | <math>10000</math>
 +
  | <math>9.357\times 10^{-11}</math>
 +
  | <math>9.357\times 10^{7}</math>
 +
  | <math>3.644\times 10^{7}</math>
 +
  | <math>3.644\times 10^{4}</math>
 +
  | <math>3.751\times 10^{5}</math>
 +
|-
 +
  | <math>5000 </math>
 +
  | <math>3.743\times 10^{-10}</math>
 +
  | <math>3.743\times 10^{8}</math>
 +
  | <math>1.458\times 10^{8}</math>
 +
  | <math>1.458\times 10^{5}</math>
 +
  | <math>1.501\times 10^{6}</math>
 +
|-
 +
  | <math>1000 </math>
 +
  | <math>9.357\times 10^{-9}</math>
 +
  | <math>9.357\times 10^{9}</math>
 +
  | <math>3.644\times 10^{9}</math>
 +
  | <math>3.644\times 10^{6}</math>
 +
  |<math>3.751\times 10^{7}</math>
 +
|-
 +
  | <math>500</math>
 +
  | <math>3.743\times 10^{-8}</math>
 +
  | <math>3.743\times 10^{10}</math>
 +
  | <math>1.458\times 10^{10}</math>
 +
  | <math>1.458\times 10^{7}</math>
 +
  | <math>1.501\times 10^{8}</math>
 +
|}
 +
 +
===CM to Lab Frame===
 +
We can substitute in for <math>\theta</math>
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin(\theta^*)sin(\theta^*)sin(\theta^*)sin(\theta^*)}</math></center>
 +
 +
 +
Using,
 +
<center><math>sin(\theta^*)=sin(\theta_{2}^*)=\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)</math></center>
 +
 +
 +
{| class="wikitable" align="center"
 +
| style="background: gray"      | <math>\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\, \partial \Omega} = \frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\,\partial \Omega^*}\frac{p}{p^{*}}</math>
 +
|}
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (3+cos^2\theta^*)^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
Now, using the trigometric identity,
 +
<center><math>sin^2 t+cos^2 t=1\Longrightarrow cos^2(\theta^*)=1-sin^2(\theta^*)</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (3+1-sin^2(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-sin(\theta^*)sin(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
Substituting,
 +
<center><math>p_{2}^*=\sqrt{E_{2}^{*2}-m^2}</math></center>
 +
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (\sqrt{E_{2}^{*2}-m^2})^4}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (E_{2}^{*2}-m^2)^2}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
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Substituting in for m, E<sub>2</sub><sup>*,</sup>and E<sup>*</sup>
 +
<math>\alpha^2=5.3279\times 10^{-5}</math>
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=(\frac{ 5.3279\times 10^{-5}( ((53.015MeV)^{2}-(.511MeV)^2)^2}{4\times (106.031MeV)^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
 +
===Substituting for Moller range and energies===
 +
Converting the number of electrons to barns,
 +
 +
 +
<center><math>\frac{d\sigma}{d\Omega_{2}'}=\frac{dN}{\mathcal L d\Omega}</math></center>
 +
 +
 +
 +
 +
<center><math>\Longrightarrow dN=\frac{d \sigma}{d \Omega} \mathcal L</math></center>
 +
 +
 +
<center><math>where \mathcal L=i\, \rho I</math></center>
 +
 +
 +
where ρ<sub>target</sub> is the density of the target material, l<sub>target</sub> is the length of the target, and N is the number of incident particles scattered.
 +
 +
 +
<center><math>\mathcal L=\frac{.86g}{1 cm^3}\times \frac{(100cm)^3}{1m^3} \times \frac{1 kg}{1000 g}\times \left[ \left(.75 \frac{1 mole}{1.01 g} \times \frac{1000g}{1 kg} \right)+\left(.25 \frac{1 mole}{14.01 g} \times \frac{1000g}{1 kg} \right)\right] \times \frac{6.022\times10^{23}particles}{1 mole} \times \frac{1cm}{100 cm} \times \frac{1 m}{ } \times \frac{10^{-28} m^2}{barn} </math></center>
 +
 +
 +
 +
<center><math>\mathcal L=\frac{860kg}{1 m^3}\times \left[ \left(\frac{742.574 mole}{kg} \right)+\left(\frac{17.844 mole}{1 kg} \right)\right] \times \frac{6.022\times 10^{-7}\ particles\cdot m^3}{1 mole\cdot barn} </math></center>
 +
 +
 +
 +
 +
<center><math>\mathcal L=\frac{5.170\times 10^{-4}kg\cdot particles}{1 mole\cdot barn} \left(\frac{760.418 mole}{kg} \right)</math></center>
 +
 +
 +
 +
<center><math>\mathcal L=\frac{.394\ particles}{1 barn} </math></center>
 +
 +
 +
 +
<center><math>\Longrightarrow N=\sigma \frac{.394\ particles}{1 barn}</math></center>
 +
 +
 +
 +
{| class="wikitable" align="center" border=1
 +
  |+ '''Number of electrons from Moller electron Momentum'''
 +
|-
 +
  ! <math>p_2^{'}(MeV/c)</math>
 +
  ! <math>\sigma(b)</math>
 +
  ! <math>Number of electrons</math>
 +
|-
 +
  | <math>\equiv 10000</math>
 +
  | <math>3.751\times 10^{5}</math>
 +
  |<math>\approx 1.478\times 10^5</math>
 +
|-
 +
  | <math>\equiv 5000 </math>
 +
  | <math>1.501\times 10^{6}</math>
 +
  |<math>\approx5.914\times 10^5</math>
 +
|-
 +
  | <math>\equiv 1000 </math>
 +
  |<math>3.751\times 10^{7}</math>
 +
  |<math>\approx 1.478\times 10^7</math>
 +
|-
 +
  | <math>\equiv 500</math>
 +
  | <math>1.501\times 10^{8}</math>
 +
  |<math>\approx 5.914\times 10^7</math>
 +
|}
 +
 +
 +
<center>[[File:Lab_Frame_Moller_DiffX.png]]</center>

Latest revision as of 19:29, 9 March 2016

Moller Differential Cross Section

Using the equation from [1]

[math]\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{*2}}\left \{\frac{1+cos^4(\frac{\theta^*}{2})}{sin^4(\frac{\theta^*}{2})}+\frac{1+sin^4(\frac{\theta^*}{2})}{cos^4(\frac{\theta^*}{2})}+\frac{2}{sin^2(\frac{\theta^*}{2})cos^2(\frac{\theta^*}{2})} \right \}[/math]


[math]where\ \alpha=\frac{e^2}{\hbar c}\quad with\quad \hbar = c =1\ and\ \theta^*=\theta^*_1=\theta^*_2[/math]


This can be simplified to the form


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}[/math]

Plugging in the values expected for 2 scattering electrons:



[math]\alpha ^2=5.3279\times 10^{-5}[/math]


[math]E^*\approx 106.031 MeV[/math]


Using unit analysis on the term outside the parantheses, we find that the differential cross section for an electron at this momentum should be around

[math]\frac{5.3279\times 10^{-5}}{4\times 1.124\times 10^{16}eV^2}=1.18\times 10^{-21} eV^{-2}=\frac{1.18\times 10^{-21}}{1eV^2}\times \frac{1\times 10^{18} }{1\times 10^{18}}=\frac{.0012}{GeV^2}[/math]

Using the conversion of


[math]\frac{1}{1GeV^2}=.3894 mb[/math]


[math]\frac{.0012}{1GeV^2}=\frac{.0012}{1}\frac{1}{1GeV^2}=.0012\times .3894 mb=.467\times 10^{-3}mb[/math]



We find that the differential cross section scale is [math]\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b[/math]

Different p21 Values

Using the conversion of


[math]\frac{1}{1GeV^2}=.3894 mb[/math]


[math]\sigma=\int d\sigma=\int \frac{d\sigma}{d\Omega_2'}d\Omega[/math]


The range of the detector is considered to be [math] .10 \le \theta \le .87[/math],[math]-\pi \le \phi \le \pi[/math]


[math]\sigma=\int_{ .611}^{2.531} \int_{-\pi}^{\pi} \frac{d\sigma}{d\Omega_2'}sin\theta \,d\theta \, d\phi [/math]


[math]\sigma=2\pi \int_{.611}^{2.531} \frac{d\sigma}{d\Omega_2'} sin\theta \,d\theta [/math]


[math]\sigma=2\pi (1.638)\frac{d\sigma}{d\Omega_2'} [/math]


[math]\sigma=(10.294) \frac{d\sigma}{d\Omega_2'} [/math]


Differential Cross Section Scale for Different p21 Values
[math]p_{2}'(MeV)[/math] [math]\frac{d\sigma}{d\Omega_{2}^'}(eV^{-2})[/math] [math]\frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2})[/math] [math]\frac{d\sigma}{d\Omega_{2}^'}(mb)[/math] [math]\frac{d\sigma}{d\Omega_{2}^'}(b)[/math] [math]\sigma(b)[/math]
[math]10000[/math] [math]9.357\times 10^{-11}[/math] [math]9.357\times 10^{7}[/math] [math]3.644\times 10^{7}[/math] [math]3.644\times 10^{4}[/math] [math]3.751\times 10^{5}[/math]
[math]5000 [/math] [math]3.743\times 10^{-10}[/math] [math]3.743\times 10^{8}[/math] [math]1.458\times 10^{8}[/math] [math]1.458\times 10^{5}[/math] [math]1.501\times 10^{6}[/math]
[math]1000 [/math] [math]9.357\times 10^{-9}[/math] [math]9.357\times 10^{9}[/math] [math]3.644\times 10^{9}[/math] [math]3.644\times 10^{6}[/math] [math]3.751\times 10^{7}[/math]
[math]500[/math] [math]3.743\times 10^{-8}[/math] [math]3.743\times 10^{10}[/math] [math]1.458\times 10^{10}[/math] [math]1.458\times 10^{7}[/math] [math]1.501\times 10^{8}[/math]

CM to Lab Frame

We can substitute in for [math]\theta[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin(\theta^*)sin(\theta^*)sin(\theta^*)sin(\theta^*)}[/math]


Using,

[math]sin(\theta^*)=sin(\theta_{2}^*)=\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)[/math]


[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\, \partial \Omega} = \frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\,\partial \Omega^*}\frac{p}{p^{*}}[/math]
[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)}[/math]



[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (3+cos^2\theta^*)^2}{sin^4 \left( \theta_{2}'\right)}[/math]


Now, using the trigometric identity,

[math]sin^2 t+cos^2 t=1\Longrightarrow cos^2(\theta^*)=1-sin^2(\theta^*)[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (3+1-sin^2(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-sin(\theta^*)sin(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}[/math]


Substituting,

[math]p_{2}^*=\sqrt{E_{2}^{*2}-m^2}[/math]



[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (\sqrt{E_{2}^{*2}-m^2})^4}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (E_{2}^{*2}-m^2)^2}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}[/math]


Substituting in for m, E2*,and E* [math]\alpha^2=5.3279\times 10^{-5}[/math]

[math]\frac{d\sigma}{d\Omega '_1}=(\frac{ 5.3279\times 10^{-5}( ((53.015MeV)^{2}-(.511MeV)^2)^2}{4\times (106.031MeV)^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}[/math]


[math]\frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}[/math]

Substituting for Moller range and energies

Converting the number of electrons to barns,


[math]\frac{d\sigma}{d\Omega_{2}'}=\frac{dN}{\mathcal L d\Omega}[/math]



[math]\Longrightarrow dN=\frac{d \sigma}{d \Omega} \mathcal L[/math]


[math]where \mathcal L=i\, \rho I[/math]


where ρtarget is the density of the target material, ltarget is the length of the target, and N is the number of incident particles scattered.


[math]\mathcal L=\frac{.86g}{1 cm^3}\times \frac{(100cm)^3}{1m^3} \times \frac{1 kg}{1000 g}\times \left[ \left(.75 \frac{1 mole}{1.01 g} \times \frac{1000g}{1 kg} \right)+\left(.25 \frac{1 mole}{14.01 g} \times \frac{1000g}{1 kg} \right)\right] \times \frac{6.022\times10^{23}particles}{1 mole} \times \frac{1cm}{100 cm} \times \frac{1 m}{ } \times \frac{10^{-28} m^2}{barn} [/math]


[math]\mathcal L=\frac{860kg}{1 m^3}\times \left[ \left(\frac{742.574 mole}{kg} \right)+\left(\frac{17.844 mole}{1 kg} \right)\right] \times \frac{6.022\times 10^{-7}\ particles\cdot m^3}{1 mole\cdot barn} [/math]



[math]\mathcal L=\frac{5.170\times 10^{-4}kg\cdot particles}{1 mole\cdot barn} \left(\frac{760.418 mole}{kg} \right)[/math]


[math]\mathcal L=\frac{.394\ particles}{1 barn} [/math]


[math]\Longrightarrow N=\sigma \frac{.394\ particles}{1 barn}[/math]


Number of electrons from Moller electron Momentum
[math]p_2^{'}(MeV/c)[/math] [math]\sigma(b)[/math] [math]Number of electrons[/math]
[math]\equiv 10000[/math] [math]3.751\times 10^{5}[/math] [math]\approx 1.478\times 10^5[/math]
[math]\equiv 5000 [/math] [math]1.501\times 10^{6}[/math] [math]\approx5.914\times 10^5[/math]
[math]\equiv 1000 [/math] [math]3.751\times 10^{7}[/math] [math]\approx 1.478\times 10^7[/math]
[math]\equiv 500[/math] [math]1.501\times 10^{8}[/math] [math]\approx 5.914\times 10^7[/math]


Lab Frame Moller DiffX.png