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Rad Data Handbook 20.. - Voss Associates

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NEUTRON SHIELD THICKNESS<br />

I = I0e -Nx<br />

where; I = final neutron flux rate<br />

I 0 = initial neutron flux rate<br />

= shield cross section in square centimeters<br />

N<br />

3<br />

= number of atoms per cm in the shield<br />

x = shield thickness in centimeters<br />

example:<br />

A dosimetry phantom is designed to simulate the<br />

composition of the human body. Ten % by weight is<br />

hydrogen. Assume a density of 1 and a shield cross<br />

section of hydrogen of 0.1 barns. A barn is 1E-24<br />

2 3<br />

cm . N, the number of atoms per cm , is 10% of<br />

Avogadro’s number, so N equals 6E22 hydrogen<br />

3<br />

atoms per cm . Assume the phantom thickness is<br />

30 cm.<br />

2<br />

I 0 = 5,000 n/cm /s<br />

2<br />

= 1E-25 cm (0.1 barns)<br />

N = 6E22 atoms per cm 3<br />

x = 30 centimeters thick<br />

-Nx = 1E-25 times 6E22 times 30 = -0.18<br />

I = I0e -Nx<br />

2 -0.18<br />

I = 5,000 n/cm /s e<br />

2 2<br />

I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s<br />

The attenuation of the neutron flux by the phantom is<br />

about 16%.<br />

59<br />

NEUTRON SHIELD THICKNESS<br />

I = I0e -Nx<br />

where; I = final neutron flux rate<br />

I 0 = initial neutron flux rate<br />

= shield cross section in square centimeters<br />

N<br />

3<br />

= number of atoms per cm in the shield<br />

x = shield thickness in centimeters<br />

example:<br />

A dosimetry phantom is designed to simulate the<br />

composition of the human body. Ten % by weight is<br />

hydrogen. Assume a density of 1 and a shield cross<br />

section of hydrogen of 0.1 barns. A barn is 1E-24<br />

2 3<br />

cm . N, the number of atoms per cm , is 10% of<br />

Avogadro’s number, so N equals 6E22 hydrogen<br />

3<br />

atoms per cm . Assume the phantom thickness is<br />

30 cm.<br />

2<br />

I 0 = 5,000 n/cm /s<br />

2<br />

= 1E-25 cm (0.1 barns)<br />

N = 6E22 atoms per cm 3<br />

x = 30 centimeters thick<br />

-Nx = 1E-25 times 6E22 times 30 = -0.18<br />

I = I0e -Nx<br />

2 -0.18<br />

I = 5,000 n/cm /s e<br />

2 2<br />

I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s<br />

The attenuation of the neutron flux by the phantom is<br />

about 16%.<br />

59<br />

NEUTRON SHIELD THICKNESS<br />

I = I0e -Nx<br />

where; I = final neutron flux rate<br />

I 0 = initial neutron flux rate<br />

= shield cross section in square centimeters<br />

N<br />

3<br />

= number of atoms per cm in the shield<br />

x = shield thickness in centimeters<br />

example:<br />

A dosimetry phantom is designed to simulate the<br />

composition of the human body. Ten % by weight is<br />

hydrogen. Assume a density of 1 and a shield cross<br />

section of hydrogen of 0.1 barns. A barn is 1E-24<br />

2 3<br />

cm . N, the number of atoms per cm , is 10% of<br />

Avogadro’s number, so N equals 6E22 hydrogen<br />

3<br />

atoms per cm . Assume the phantom thickness is<br />

30 cm.<br />

2<br />

I 0 = 5,000 n/cm /s<br />

2<br />

= 1E-25 cm (0.1 barns)<br />

N = 6E22 atoms per cm 3<br />

x = 30 centimeters thick<br />

-Nx = 1E-25 times 6E22 times 30 = -0.18<br />

I = I0e -Nx<br />

2 -0.18<br />

I = 5,000 n/cm /s e<br />

2 2<br />

I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s<br />

The attenuation of the neutron flux by the phantom is<br />

about 16%.<br />

59<br />

NEUTRON SHIELD THICKNESS<br />

I = I0e -Nx<br />

where; I = final neutron flux rate<br />

I 0 = initial neutron flux rate<br />

= shield cross section in square centimeters<br />

N<br />

3<br />

= number of atoms per cm in the shield<br />

x = shield thickness in centimeters<br />

example:<br />

A dosimetry phantom is designed to simulate the<br />

composition of the human body. Ten % by weight is<br />

hydrogen. Assume a density of 1 and a shield cross<br />

section of hydrogen of 0.1 barns. A barn is 1E-24<br />

2 3<br />

cm . N, the number of atoms per cm , is 10% of<br />

Avogadro’s number, so N equals 6E22 hydrogen<br />

3<br />

atoms per cm . Assume the phantom thickness is<br />

30 cm.<br />

2<br />

I 0 = 5,000 n/cm /s<br />

2<br />

= 1E-25 cm (0.1 barns)<br />

N = 6E22 atoms per cm 3<br />

x = 30 centimeters thick<br />

-Nx = 1E-25 times 6E22 times 30 = -0.18<br />

I = I0e -Nx<br />

2 -0.18<br />

I = 5,000 n/cm /s e<br />

2 2<br />

I = 5,000 n/cm /s x 0.835 = 4,175 n/cm /s<br />

The attenuation of the neutron flux by the phantom is<br />

about 16%.<br />

59

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