Difference between revisions of "Counts Rate (44 MeV LINAC)"
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=number of electrons/sec on radiator= | =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><br><br> | <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><br><br> | ||
| Line 19: | Line 18: | ||
radiation length of Ti(Z=22)is 3.59 cm<br> | radiation length of Ti(Z=22)is 3.59 cm<br> | ||
Assume the radiator thickness is <math>12.5 \mu m</math><br><br> | Assume the radiator thickness is <math>12.5 \mu m</math><br><br> | ||
| − | |||
<math>3.59 cm/12.5\mu m = 3.48 \cdot 10^{-4} \ r.l.</math><br> | <math>3.59 cm/12.5\mu m = 3.48 \cdot 10^{-4} \ r.l.</math><br> | ||
| − | + | ==steps together== | |
| − | == | + | <math>o.1 \frac{\gamma 's}{(e^-\\ MeV\ r.l.)} \times 3.48 \cdot 10^{-4} r.l. \times 10\ MeV \times 0.47 \cdot 10^{13} \frac{e^-}{sec}=1.63 \cdot 10^{9} \frac{\gamma}{sec}</math><br><br> |
| − | <math> | ||
=number of neutrons in 1 second= | =number of neutrons in 1 second= | ||
=geometrical factor, isotropic case= | =geometrical factor, isotropic case= | ||
Revision as of 20:50, 17 May 2010
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
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
steps together