Difference between revisions of "Counts Rate (44 MeV LINAC)"

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from plot above we have about 0.1 photons/electrons/MeV/r.l from 10 MeV to 20 MeV region
 
from plot above we have about 0.1 photons/electrons/MeV/r.l from 10 MeV to 20 MeV region
 
==Radiation length==
 
==Radiation length==
radiation length of Ti(Z=22)is 3.59 cm<>     
+
radiation length of Ti(Z=22)is 3.59 cm<br>     
 
Assume the radiator thickness is <math>12.5 \mu m</math><br>
 
Assume the radiator thickness is <math>12.5 \mu m</math><br>
 
Which corresponds to <math>3.59 cm/12.5\mu m = 3.48 /cdot 10^{-4} \ r.l.</math><br>
 
Which corresponds to <math>3.59 cm/12.5\mu m = 3.48 /cdot 10^{-4} \ r.l.</math><br>

Revision as of 20:43, 17 May 2010

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LINAC parameters used in calculations

1) pulse width 50 ns
2) pulse current 50 A
3) repetition rate 300 Hz
4) energy 44 MeV

number of electrons/sec on radiator

[math] 50ps \times 50A \times 300Hz \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}C} = 0.47 \cdot 10^{13} \frac{e^-}{sec}[/math]

number of photons/sec from radiator

Bremsstrahlung

from plot above we have about 0.1 photons/electrons/MeV/r.l from 10 MeV to 20 MeV region

Radiation length

radiation length of Ti(Z=22)is 3.59 cm
Assume the radiator thickness is [math]12.5 \mu m[/math]
Which corresponds to [math]3.59 cm/12.5\mu m = 3.48 /cdot 10^{-4} \ r.l.[/math]


[math]10^{-3} \frac{\gamma 's}{(e^-\ MeV\ r.l.)} \times 2 \cdot 10^{-4} r.l. \times 44 MeV \times 0.375 \cdot 10^{12} \frac{e^-}{sec}=0.3 \cdot 10^{7} \frac{\gamma}{sec}[/math]

number of neutrons in 1 second

geometrical factor, isotropic case