2-Neutron Correlation

Big Detector Solid Angle Calculations

MCNPX Simulation

• 14 MeV neutron source, emitted isotropically ([math]4\pi[/math])
  
• Detector placed 1m away from source
  

• face of the detector is 15.24cm x 76.2cm, and 3.6cm deep
  

The solid angle can be found from the number of particles hitting the detector as:
[math]\Delta \Omega = 4\pi*\frac{hits}{hits + misses}[/math]

Results

• Out of 1E9 neutrons generated, 8618287 neutrons hit the detector
  • [math]\Delta \Omega = 0.108 Sr[/math]
    • if the detector is placed 70cm away from the source, [math]\Delta \Omega = 0.207 Sr[/math]
    • if the detector is placed 65cm away from the source, [math]\Delta \Omega = 0.236 Sr[/math]

• As a test to verify our results
  • We change the detector size to 2cm by 2cm and used 1E9 neutrons again
  • 32061 neutrons struck the detector
  • [math]\Delta \Omega = 0.0004 Sr[/math]

• And, as a second test to verify our results
  • We change the detector size to 1cm by 1cm and used 1E9 neutrons again
  • 7965 neutrons struck the detector
  • [math]\Delta \Omega = 0.0001 Sr[/math]

Now, what neutron singles rate into the detector should correspond to 1 fission per pulse?

• If we have 1 fission per pulse and each fission emits on average 2.3 neutrons, we should expect 2.3 neutrons/pulse
• The number of neutrons hitting the detector per pulse is found as [math]2.3*\frac{\Delta \Omega}{4\pi}[/math]
  • @ 1 meter => 0.0198 neutrons hitting the detector per pulse
  • @ 70 cm => 0.0379 neutrons hitting the detector per pulse
• Taking into account the efficiency of the detector [math]\epsilon_0[/math], the number detected per pulse can be found as [math]2.3*\frac{\Delta \Omega}{4\pi}*\epsilon_0[/math]
  • @ 1 meter from source => ([math]0.0198*\epsilon_0=0.0198*0.17=0.003[/math]) neutrons detected per pulse
  • @ 70 cm from source => ([math]0.0379*\epsilon_0=0.0379*0.17=0.006[/math]) neutrons detected per pulse

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