What Fraction of Protons Interact in a Liquid Hydrogen Target?

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SUMMARY

The discussion focuses on calculating the fraction of protons that interact with a liquid hydrogen target when subjected to a proton beam of energy E=7mpc². The volume of the target is specified as V=20 cm³ with a density of ρ=0.071 g cm⁻³ and a beam spot area of A=4 cm². The total cross section is given as σ=30 mb, leading to a calculated interaction fraction of f=6.4E-6 and a mean free path of λ=7.84 km. The calculations were derived using the equations for number density and mean free path, but the results raised concerns regarding their order of magnitude.

PREREQUISITES
  • Understanding of particle physics concepts, specifically proton interactions.
  • Familiarity with the equations for mean free path and number density.
  • Knowledge of liquid hydrogen properties, including density and cross section.
  • Basic proficiency in dimensional analysis and unit conversions.
NEXT STEPS
  • Review the derivation of the mean free path formula λ=1/nσ.
  • Explore the implications of cross section measurements in particle physics.
  • Investigate the properties of liquid hydrogen as a target material in high-energy physics experiments.
  • Learn about the statistical methods used in analyzing particle interactions and collision probabilities.
USEFUL FOR

This discussion is beneficial for physics students, researchers in particle physics, and anyone involved in experimental design using liquid hydrogen targets for proton beam interactions.

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Homework Statement


A proton beam of energy E=7mpc2 is incident on a cylindrical volume (V=20 cm3) of liquid hydrogen (density ρ=0.071 g cm-3) at rest. The spot area of the beam is A=4 cm2 and the total cross section is σ=30 mb. What fraction of the protons entering the target will interact? What is the mean free path (collision length)?

Homework Equations


λ=1/nσ

The Attempt at a Solution


I found the number of protons in the given volume, N = ρV/mp. I then computed the fraction as f=Nσ /A=ρVσ/mA=6.4E-6. I don't like the order of magnitude I am getting from this approach. I found the mean free path from the equation above. Again, with a weird order of magnitude. λ=m/ρσ=7.84 km.
 
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