NEUTRON SHIELD THICKNESS
0
I
=
I e
-
F
Nx
where;
I
= final neutron flux rate
0
I
= initial neutron flux rate
F
= shield cross section in square centim eters
N = num ber of atom s per cm in the shield
3
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
cm . N, the num ber of atom s per cm , is 10% of
2
3
Avogadro’s num ber, so N equals 6E22 hydrogen
atom s per cm . Assum e 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 atom s per cm
3
x
= 30 centim eters thick
-
F
Nx = 1E-25 tim es 6E22 tim es 30 = -0.18
0
I
=
I e
-
F
Nx
I
=
5,000 n/cm /s e
2
-0.18
I
=
5,000 n/cm /s x 0.835 = 4,175 n/cm /s
2
2
The attenuation of the neutron flux by the phantom is
about 16% .
59
NEUTRON SHIELD THICKNESS
0
I
=
I e
-
F
Nx
where;
I
= final neutron flux rate
0
I
= initial neutron flux rate
F
= shield cross section in square centim eters
N = num ber of atom s per cm in the shield
3
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
cm . N, the num ber of atom s per cm , is 10% of
2
3
Avogadro’s num ber, so N equals 6E22 hydrogen
atom s per cm . Assum e 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 atom s per cm
3
x
= 30 centim eters thick
-
F
Nx = 1E-25 tim es 6E22 tim es 30 = -0.18
0
I
=
I e
-
F
Nx
I
=
5,000 n/cm /s e
2
-0.18
I
=
5,000 n/cm /s x 0.835 = 4,175 n/cm /s
2
2
The attenuation of the neutron flux by the phantom is
about 16% .
59
NEUTRON SHIELD THICKNESS
0
I
=
I e
-
F
Nx
where;
I
= final neutron flux rate
0
I
= initial neutron flux rate
F
= shield cross section in square centim eters
N = num ber of atom s per cm in the shield
3
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
cm . N, the num ber of atom s per cm , is 10% of
2
3
Avogadro’s num ber, so N equals 6E22 hydrogen
atom s per cm . Assum e 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 atom s per cm
3
x
= 30 centim eters thick
-
F
Nx = 1E-25 tim es 6E22 tim es 30 = -0.18
0
I
=
I e
-
F
Nx
I
=
5,000 n/cm /s e
2
-0.18
I
=
5,000 n/cm /s x 0.835 = 4,175 n/cm /s
2
2
The attenuation of the neutron flux by the phantom is
about 16% .
59
NEUTRON SHIELD THICKNESS
0
I
=
I e
-
F
Nx
where;
I
= final neutron flux rate
0
I
= initial neutron flux rate
F
= shield cross section in square centim eters
N = num ber of atom s per cm in the shield
3
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
cm . N, the num ber of atom s per cm , is 10% of
2
3
Avogadro’s num ber, so N equals 6E22 hydrogen
atom s per cm . Assum e 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 atom s per cm
3
x
= 30 centim eters thick
-
F
Nx = 1E-25 tim es 6E22 tim es 30 = -0.18
0
I
=
I e
-
F
Nx
I
=
5,000 n/cm /s e
2
-0.18
I
=
5,000 n/cm /s x 0.835 = 4,175 n/cm /s
2
2
The attenuation of the neutron flux by the phantom is
about 16% .
59