Aluminum Converter

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: $f=300Hz$

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.

$N = \frac {f I ∆t} {e} = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 4.6875*10^{12}$

So, we have around $4.6875*10^{12}$ electrons per second or $15.625*10^9$ 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 $1.78 MeV cm^2/g$.

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

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

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