Revision as of 19:59, 7 June 2010

Calculating the temperature of a 1/2 mil Aluminum converter with energy deposited from a 44 MeV electron beam.

We have electron beam of:

Frequency: $f=300Hz$

Peak current: $I=50 Amps$

Pulse width: $∆t= 50 ps = 5*10^{-11} seconds$

By $Q=It$ , we have $N*e=f*I*∆t$

Where $N$ is the number of electrons that hit the target per second, $e$ is electron charge and $f$ , $I$ and $∆t$ are given above.

So, we have around $4.6875*10^{12}$ electrons per second or $15.625*10^9$ electrons per pulse.

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

$1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm$

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

@@ Line 15: / Line 15: @@
 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>
+<math> N = \frac {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.
@@ Line 28: / Line 28: @@
-Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A=19.62 mm^2 <\math>.
+Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A = 19.62 mm^2 </math>.