Rad Data Handbook 20.. - Voss Associates

Neutron Half-Value Layers in centimeters
Energy in MeV 1 5 10 15
Polyethylene 3.7 6.1 7.7 8.8
W ater 4.3 6.9 8.8 10.1
Concrete 6.8 11 14 16
Damp soil 8.8 14.3 18.2 20.8
example:
How many half-value layers of polyethylene are needed
to attenuate a 100 mRem/hr 5 MeV neutron source to 5
mRem/hr? How thick does the polyehylene need to be?
I = I 0 x 0.5 n
I = 5 mRem/hr
I 0 = 100 mRem/hr
n = the number of half-value layers
I/I 0 = 0.5 n
5/100 = 0.05 = 0.5 n
ln 0.05 = n x ln 0.5
ln 0.05/ln 0.5 = n
-2.996/-0.693 = n
4.32 = n
It will take 4.32 half-value layers of polyethylene to
reduce attenuate the neutron source.
4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm
Neutron Half-Value Layers in centimeters
Energy in MeV 1 5 10 15
Polyethylene 3.7 6.1 7.7 8.8
W ater 4.3 6.9 8.8 10.1
Concrete 6.8 11 14 16
Damp soil 8.8 14.3 18.2 20.8
example:
How many half-value layers of polyethylene are needed
to attenuate a 100 mRem/hr 5 MeV neutron source to 5
mRem/hr? How thick does the polyehylene need to be?
I = I 0 x 0.5 n
I = 5 mRem/hr
I 0 = 100 mRem/hr
n = the number of half-value layers
I/I 0 = 0.5 n
5/100 = 0.05 = 0.5 n
ln 0.05 = n x ln 0.5
ln 0.05/ln 0.5 = n
-2.996/-0.693 = n
4.32 = n
It will take 4.32 half-value layers of polyethylene to
reduce attenuate the neutron source.
4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm
60
60
Neutron Half-Value Layers in centimeters
Energy in MeV 1 5 10 15
Polyethylene 3.7 6.1 7.7 8.8
W ater 4.3 6.9 8.8 10.1
Concrete 6.8 11 14 16
Damp soil 8.8 14.3 18.2 20.8
example:
How many half-value layers of polyethylene are needed
to attenuate a 100 mRem/hr 5 MeV neutron source to 5
mRem/hr? How thick does the polyehylene need to be?
I = I 0 x 0.5 n
I = 5 mRem/hr
I 0 = 100 mRem/hr
n = the number of half-value layers
I/I 0 = 0.5 n
5/100 = 0.05 = 0.5 n
ln 0.05 = n x ln 0.5
ln 0.05/ln 0.5 = n
-2.996/-0.693 = n
4.32 = n
It will take 4.32 half-value layers of polyethylene to
reduce attenuate the neutron source.
4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm
Neutron Half-Value Layers in centimeters
Energy in MeV 1 5 10 15
Polyethylene 3.7 6.1 7.7 8.8
W ater 4.3 6.9 8.8 10.1
Concrete 6.8 11 14 16
Damp soil 8.8 14.3 18.2 20.8
example:
How many half-value layers of polyethylene are needed
to attenuate a 100 mRem/hr 5 MeV neutron source to 5
mRem/hr? How thick does the polyehylene need to be?
I = I 0 x 0.5 n
I = 5 mRem/hr
I 0 = 100 mRem/hr
n = the number of half-value layers
I/I 0 = 0.5 n
5/100 = 0.05 = 0.5 n
ln 0.05 = n x ln 0.5
ln 0.05/ln 0.5 = n
-2.996/-0.693 = n
4.32 = n
It will take 4.32 half-value layers of polyethylene to
reduce attenuate the neutron source.
4.32 half-value layers is 4.32 x 6.1 cm = 26.4 cm
60
60