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 = \frac {300*50*5*10^{-11}} {1.6*10^{-19}} = 4.6875*10^{12} </math>
+
<math> N = f*I*∆t/e = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 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.
 
So, we have around <math> 4.6875*10^{12}</math> electrons per second or <math> 15.625*10^9 </math> electrons per pulse.
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From NIST ([http://physics.nist.gov/cgi-bin/Star/e_table.pl] 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>.  
 
From NIST ([http://physics.nist.gov/cgi-bin/Star/e_table.pl] 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>
+
<math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>
  
  

Revision as of 19:57, 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 = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 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 = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 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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