What is the expected rate of reactions in the detector?

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SUMMARY

The expected rate of reactions in a neutrino detector can be calculated using the formula W = σ J n δx. In this scenario, the reactor emits 1021 neutrinos per second, and the detector contains a liquid with a density of 870 kg/m3 and a carbon-to-hydrogen ratio of 30:60. The number density of carbon atoms is calculated to be 3.5 x 1029 m-3. The cross section for the reaction is σ = 10-46 m2, and the distance from the reactor to the detector is 1000 m.

PREREQUISITES
  • Understanding of neutrino interactions and cross sections
  • Familiarity with particle density calculations
  • Knowledge of the formula W = σ J n δx
  • Basic principles of flux in particle physics
NEXT STEPS
  • Calculate the expected rate of reactions using different thickness values
  • Explore the concept of flux in particle physics using the formula J/4πr2
  • Investigate the implications of varying the detector's distance from the reactor
  • Learn about the significance of cross sections in particle detection
USEFUL FOR

Students and researchers in particle physics, particularly those studying neutrino interactions and detector design, will benefit from this discussion.

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Homework Statement



A reactor fires out ##10^{21}## neutrinos per second. A detector containing ##10m^3## of liquid which contains 30 carbon atoms every 60 hydrogen atoms. The detector-reactor distance is ##1000##. The cross section for the reaction is ##\sigma = 10^{-46} m^2##, and the density of the liquid is ##870 kg m^{-3}##. Find the expected rate of reactions.

Homework Equations

The Attempt at a Solution



I know that the rate of reactions is ##W = \sigma J n \delta x## where ##J## is incoming particles per second and ##n## is number density and ##\delta x## is thickness.
Number of carbon atoms per unit volume in the detector is ##\frac{60}{90} \frac{(870)(10.3)}{(1.67 \times 10^{-27}) (10.3)} = 3.5 \times 10^{29} m^{-3}##. Can I take ##J## to be ##10^{21}##? What is the thickness in this case?
 
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Or is it better to use ##W = \sigma \frac{J}{A} (nA\delta x) = \sigma \frac{J}{A} N## where I can find the flux from using ##\frac{J}{4\pi r^2}##?
 

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