Difference between revisions of "Aluminum Converter"
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Where <math> N </math> is the number of electrons that hit the target per second, <math> e </math> is electron charge and <math> f </math>, <math> I </math> and <math> ∆t </math> are given above. | Where <math> N </math> is the number of electrons that hit the target per second, <math> e </math> is electron charge and <math> f </math>, <math> I </math> and <math> ∆t </math> are given above. | ||
− | <math> N= f*I*∆t/e=300*50*5*10^ | + | <math> N = f*I*∆t/e = \frac {300*50*5*10^{-11}} {1.6*10^{-19}} = 4.6875*10^{12} </math> |
− | So, we have around <math> 4.6875*10^ | + | So, we have around <math> 4.6875*10^{12}</math> electrons per second or <math> 15.625*10^9 </math> electrons per pulse. |
==Calculating the stopping power due to collision of one 44 MeV electron in Aluminum== | ==Calculating the stopping power due to collision of one 44 MeV electron in Aluminum== |
Revision as of 19:53, 7 June 2010
Calculating the temperature of a 1/2 mil Aluminum converter with energy deposited from a 44 MeV electron beam.
Calculating number of particles per second
We have electron beam of:
Frequency:
Peak current:
Pulse width:
By
, we haveWhere
is the number of electrons that hit the target per second, is electron charge and , and are given above.
So, we have around
electrons per second or electrons per pulse.Calculating the stopping power due to collision of one 44 MeV electron in Aluminum
From NIST ([1] see link here) the stopping power for one electron with energy of 44 MeV in Aluminum is .
Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A=19.62 mm^2 <\math>.