Difference between revisions of "Aluminum Converter"

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Peak current: <math> I=50 Amps </math>
 
Peak current: <math> I=50 Amps </math>
  
Pulse width: <math> ∆t= 50 ps = 5*10^-11 seconds </math>
+
Pulse width: <math> ∆t= 50 ps = 5*10^{-11} seconds </math>
  
 
By <math> Q=It </math>, we have <math> N*e=f*I*∆t </math>
 
By <math> Q=It </math>, we have <math> N*e=f*I*∆t </math>

Revision as of 19:49, 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: [math] f=300Hz [/math]

Peak current: [math] I=50 Amps [/math]

Pulse width: [math] ∆t= 50 ps = 5*10^{-11} seconds [/math]

By [math] Q=It [/math], we have [math] N*e=f*I*∆t [/math]

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^(-11)/(1.6*10^(-19))=4.6875*10^(12) [/math]

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

From NIST ([1] see link here) the stopping power for one electron with energy of 44 MeV in Aluminum is [math] 1.78 MeV cm^2/g [/math].

[math] 1 mil = 1/1000 inch * 2.54 cm/inch = 2.54/1000 cm =0.00254 cm [/math]



Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A=19.62 mm^2 <\math>.






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