Faraday Cup Temperature

Calculating the temperature of a Faraday Cup Rod

Number of particles per second hitting one rod

Assume electron beam parameters at Faraday Cup location are:

Frequency:  [math] f=300Hz [/math]
Peak current: [math] I=50 Amps [/math]
Pulse width: [math] ∆t= 50 ps = 5*10^{-11} seconds [/math]
Beam energy: [math] E=45 MeV [/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 \frac {MeV}{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] P = (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^4 = \frac {P}{(0.924)(2A)(\sigma)} [/math]

where [math] \sigma [/math] is the Stefan-Boltzmann constant, [math] \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} [/math]. Assume a beam spot diameter on the converter surface of 5mm, or an area of [math] A = 19.62 mm^2 = 19.62*10^{-6} m^2 [/math].

Plugging in the numbers we see that the temperature will increase [math] 217.2 K [/math]. Now, adding in the temperature of the converter at room temperature we get :

[math] T = 300 + 217.2 = 517.2 K[/math]

The melting temperature of Aluminum is [math] 933.5 K [/math].

Conclusion

An Aluminum converter that is 1/2 mil thick being struck by a 44 MeV electron beam with a 50 picosecond pulse width, 300 Hz rep rate, and 50 Amp peak current is found to be safe from melting.

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