- #1

Kara386

- 208

- 2

## 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!