Radiators Temperature

1.Calculating number of particles per second

We have electron beam of:

Frequency: f=1000Hz

Peak current: I=10mAmp=0.01 Amp

Pulse width: ∆t= 50 ns=5*10-8 seconds

So, how many electrons we have in each second?

By Q=It, we have

                                            N*e=f*I*∆t

Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above. So

                          N= f*I*∆t/e=1000*0.01*5*10-8/(1.6*10-19)=3.12075*1012

So, we have around 3.12075*1012 electrons hit radiator per second.

2.Calculating Energy deposited per second

If we find the energy deposited by each electron and multiply to the total number of electrons in each second, we will find the total energy per second deposited in radiator.

To find energy deposited by each electron, we need to use formula

                                          ∆E=(dE/dx)_coll *t

Where is ∆E is energy deposited by one electron, (dE/dx)_coll is mean energy loss (also stopping power) by collision of electron and t is thickness of the radiator.

Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.

Stopping power can be found from nuclear data tables (dE/dx)ave and thickness is 0.001 times of radiation length. From Particle Data group we got radiation length and average total stopping powers around 15MeV for electrons in these materials from National Institute of Standards and Technology

Table of Radiation Lengths ( From Particle Data group. Link: [1]).

Elements	Radiation Lengths (g/cm^2)
Al	24.01
W	6.76
Ti	16.16
Fe	13.84 Table of radiator thickness, (dE/dx)_coll (from National Institute of Standards and Technology. Link: [[2]]) and energy calculations