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: [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 = \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.

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]

The effective length of 1/2 mil Al:

[math] (2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2} [/math]

The total stopping power due to collisions on Al per incident electron:

[math] (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 MeV per electron [/math]

The energy deposited per pulse:

[math] (0.0061 \frac {MeV}{electron})(15.625*10^9 \frac {electrons}{pulse}) = 95.3125*10^6 \frac {MeV}{pulse} [/math]

The energy deposited per second:

[math] (95.3125*10^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.6*10^9 \frac {MeV}{second} [/math]

Calculating the temperature increase

The power deposited in 1/2 mil Al is:

[math] 28.6*10^9 \frac {MeV}{second} = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV} = 4.575*10^{-3} W [/math]

Stefan-Boltzmann Law (Wien Approximation) says:

[math] P = (0.924)(Area)(\sigma)(T^4) [/math]

Solving for Temperature and taking into account the two sides of the converter we get:

[math] T = (\frac {P}{(0.924)(2A)(\sigma)})^{\frac {1}{4}} [/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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