Panasonic PT-DZ12000U Operating Instructions Manual page 70

NEUTRON SHIELD THICKNESS
I = I e
0
where; I
I = initial neutron flux rate
0
= shield cross section in square centim eters
N = num ber of atom s per cm in the shield
x = shield thickness in centim eters
exam ple:
A dosim etry phantom is designed to sim ulate the
com position of the hum an body. Ten % by weight is
hydrogen. Assum e a density of 1 and a shield cross
section of hydrogen of 0.1 barns. A barn is 1E-24
2
cm . N, the num ber of atom s per cm , is 10% of
Avogadro's num ber, so N equals 6E22 hydrogen
atom s per cm . Assum e the phantom thickness is
30 cm .
I = 5,000 n/cm /s
0
= 1E-25 cm (0.1 barns)
N = 6E22 atom s per cm
x = 30 centim eters thick
- Nx = 1E-25 tim es 6E22 tim es 30 = -0.18
I = I e
0
I = 5,000 n/cm /s e
I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s
The attenuation of the neutron flux by the phantom is
about 16%.
- Nx
= final neutron flux rate
3
2
2
- Nx
2
2
3
3
-0.18
59
3
2