Panasonic PT-DZ12000U - Page 70

NEUTRON SHIELD THICKNESS
0
I=Ie
-FNx
where; I = final neutron flux rate
0
I = initial neutron flux rate
F = shield cross section in square centimeters
N = number of atoms per cm in the shield
3
x = shield thickness in centimeters
example:
A dosimetry phantom is designed to simulate the
composition of the human body. Ten % by weight is
hydrogen. Assume a density of 1 and a shield cross
section of hydrogen of 0.1 barns. A barn is 1E-24
cm . N, the number of atoms per cm , is 10% of
23
Avogadro’s number, so N equals 6E22 hydrogen
atoms per cm . Assume the phantom thickness is
3
30 cm.
0
I = 5,000 n/cm /s
2
F = 1E-25 cm (0.1 barns)
2
N = 6E22 atoms per cm
3
x = 30 centimeters thick
-FNx = 1E-25 times 6E22 times 30 = -0.18
0
I=Ie
-FNx
I = 5,000 n/cm /s e
2 -0.18
I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s
22
The attenuation of the neutron flux by the phantom is
about 16%.
59
NEUTRON SHIELD THICKNESS
0
I=Ie
-FNx
where; I = final neutron flux rate
0
I = initial neutron flux rate
F = shield cross section in square centimeters
N = number of atoms per cm in the shield
3
x = shield thickness in centimeters
example:
A dosimetry phantom is designed to simulate the
composition of the human body. Ten % by weight is
hydrogen. Assume a density of 1 and a shield cross
section of hydrogen of 0.1 barns. A barn is 1E-24
cm . N, the number of atoms per cm , is 10% of
23
Avogadro’s number, so N equals 6E22 hydrogen
atoms per cm . Assume the phantom thickness is
3
30 cm.
0
I = 5,000 n/cm /s
2
F = 1E-25 cm (0.1 barns)
2
N = 6E22 atoms per cm
3
x = 30 centimeters thick
-FNx = 1E-25 times 6E22 times 30 = -0.18
0
I=Ie
-FNx
I = 5,000 n/cm /s e
2 -0.18
I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s
22
The attenuation of the neutron flux by the phantom is
about 16%.
5
9
NEUTRON SHIELD THICKNESS
0
I=Ie
-FNx
where; I = final neutron flux rate
0
I = initial neutron flux rate
F = shield cross section in square centimeters
N = number of atoms per cm in the shield
3
x = shield thickness in centimeters
example:
A dosimetry phantom is designed to simulate the
composition of the human body. Ten % by weight is
hydrogen. Assume a density of 1 and a shield cross
section of hydrogen of 0.1 barns. A barn is 1E-24
cm . N, the number of atoms per cm , is 10% of
23
Avogadro’s number, so N equals 6E22 hydrogen
atoms per cm . Assume the phantom thickness is
3
30 cm.
0
I = 5,000 n/cm /s
2
F = 1E-25 cm (0.1 barns)
2
N = 6E22 atoms per cm
3
x = 30 centimeters thick
-FNx = 1E-25 times 6E22 times 30 = -0.18
0
I=Ie
-FNx
I = 5,000 n/cm /s e
2 -0.18
I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s
22
The attenuation of the neutron flux by the phantom is
about 16%.
59
NEUTRON SHIELD THICKNESS
0
I=Ie
-FNx
where; I = final neutron flux rate
0
I = initial neutron flux rate
F = shield cross section in square centimeters
N = number of atoms per cm in the shield
3
x = shield thickness in centimeters
example:
A dosimetry phantom is designed to simulate the
composition of the human body. Ten % by weight is
hydrogen. Assume a density of 1 and a shield cross
section of hydrogen of 0.1 barns. A barn is 1E-24
cm . N, the number of atoms per cm , is 10% of
23
Avogadro’s number, so N equals 6E22 hydrogen
atoms per cm . Assume the phantom thickness is
3
30 cm.
0
I = 5,000 n/cm /s
2
F = 1E-25 cm (0.1 barns)
2
N = 6E22 atoms per cm
3
x = 30 centimeters thick
-FNx = 1E-25 times 6E22 times 30 = -0.18
0
I=Ie
-FNx
I = 5,000 n/cm /s e
2 -0.18
I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s
22
The attenuation of the neutron flux by the phantom is
about 16%.
59