Electromagnetic cascade in a calorimeter

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SUMMARY

The discussion focuses on estimating the thickness of a calorimeter required to fully contain an electromagnetic shower initiated by a 10 GeV electron. The calorimeter consists of alternating layers of lead (1.75 mm thick) and scintillator, with the radiation length of lead being 0.64 cm and the critical energy for lead at 9.6 MeV. The key equation derived for energy loss is E = E0 / 2^(t/X0), where t represents the thickness of lead. The participants clarify the structure of the calorimeter and correct the initial misunderstanding regarding the inclusion of particle doubling in the calculations.

PREREQUISITES
  • Understanding of electromagnetic showers and their behavior in calorimeters
  • Familiarity with radiation length and critical energy concepts
  • Basic knowledge of scintillator and lead properties in particle physics
  • Ability to manipulate exponential equations related to energy loss
NEXT STEPS
  • Research the principles of electromagnetic showers in particle physics
  • Study the calculation of radiation lengths in various materials
  • Learn about the design and function of calorimeters in high-energy physics experiments
  • Explore the effects of different scintillator materials on shower containment
USEFUL FOR

Students and researchers in particle physics, particularly those involved in calorimeter design and analysis of electromagnetic interactions in high-energy experiments.

Kara386
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Homework Statement


A calorimeter is made from layers of lead (1.75mm thick) alternated with layers of scintillator. The radiation length ##X_0## of lead is ##0.64cm##.

In an EM shower the number of particles doubles and the energy of each particle halves per radiation length travelled. The shower stops when critical energy ##E_c## is reached. For lead ##E_c## is 9.6MeV. Estimate the calorimeter thickness required to completely contain a shower caused by a 10GeV electron. Neglect interactions in the scintillator.

Homework Equations

The Attempt at a Solution


I know a calorimeter has to have scintillator as the first and last layers. So if there are n layers of scintillator, there will be n-1 layers of lead.

Based on the information given. I'm thinking the equation should be something like
##E = \frac{E_0}{2^{t/X_0}}##
Where t is the thickness of lead the shower travels through. Then thickness would be ##(0.175t) \times 0.4(t+1)##. Is that ok? Or do I need to somehow include the doubling in particle number in there?

Thanks for any help!
 
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Kara386 said:
I know a calorimeter has to have scintillator as the first and last layers.
It does not have to. Is there a scintillator thickness given? Otherwise you can just calculate the required length of lead.

Your second expression grows quadratically with t, that cannot be right. Did you mean "+"? Where does the 0.4 come from?
 

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