Effects Due to Target Length

From New IAC Wiki
Revision as of 16:29, 30 August 2018 by Vanwdani (talk | contribs)
Jump to navigation Jump to search

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.png

Moller Electron Angle Theta in Lab Frame
Figure 1: GEANT4 Simulation for the Moller electron scattering angle theta distribution for 6E6 incident 11 GeV electrons in the Lab frame of reference. The LH2 target reduces the mean free path of the Moller electron to around 60 degrees in the lab frame. This property of limiting the Moller angle is independent of the target length.


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]