Difference between revisions of "Effects Due to Target Length"

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As the target material length is increased, the amount of multiple scattering increases.  Since the majority of particles created in the CM frame are at large angles, this implies  
 
As the target material length is increased, the amount of multiple scattering increases.  Since the majority of particles created in the CM frame are at large angles, this implies  
  
[[File:CompareTargetLength.png]]
+
[[File:CompareTargetLength.png]][[File:MolThetaLabTargetLength.png]]
  
 
This gives, for LH2 in a 5cm long target:
 
This gives, for LH2 in a 5cm long target:

Revision as of 04:40, 28 August 2018

As the target material length is increased, the amount of multiple scattering increases. Since the majority of particles created in the CM frame are at large angles, this implies

CompareTargetLength.pngMolThetaLabTargetLength.png

This gives, for LH2 in a 5cm long target:

[math]\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =1.05\times 10^{-1} barns^{-1}[/math]


Using the number of incident electrons, for 1 Moller electron:

[math]\frac{1}{\rho_{target}\times l_{target} \times 6\times 10^6}=1.58\times 10^{-6} barns[/math]

This gives, for LH2 in a 1cm long target:

[math]\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{1 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =2.10\times 10^{-2} barns^{-1}[/math]


Using the number of incident electrons, for 1 Moller electron:

[math]\frac{1}{\rho_{target}\times l_{target} \times 6\times 10^6}=7.94\times 10^{-6} barns[/math]


This gives, for LH2 in a 1mm long target:

[math]\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{0.1 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =2.10\times 10^{-3} barns^{-1}[/math]


Using the number of incident electrons, for 1 Moller electron:

[math]\frac{1}{\rho_{target}\times l_{target} \times 6\times 10^6}=7.94\times 10^{-5} barns[/math]