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
Dam p soil
8.8
14.3
18.2
20.8
exam ple:
How m any half-value layers of polyethylene are needed
to attenuate a 100 m Rem /hr 5 MeV neutron source to 5
m Rem /hr? How thick does the polyehylene need to be?
0
I
=
I x 0.5
n
I
= 5 m Rem /hr
0
I
=
100 m Rem /hr
n
=
the num ber of half-value layers
0
I/I =
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
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
Dam p soil
8.8
14.3
18.2
20.8
exam ple:
How m any half-value layers of polyethylene are needed
to attenuate a 100 m Rem /hr 5 MeV neutron source to 5
m Rem /hr? How thick does the polyehylene need to be?
0
I
=
I x 0.5
n
I
= 5 m Rem /hr
0
I
=
100 m Rem /hr
n
=
the num ber of half-value layers
0
I/I =
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
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
Dam p soil
8.8
14.3
18.2
20.8
exam ple:
How m any half-value layers of polyethylene are needed
to attenuate a 100 m Rem /hr 5 MeV neutron source to 5
m Rem /hr? How thick does the polyehylene need to be?
0
I
=
I x 0.5
n
I
= 5 m Rem /hr
0
I
=
100 m Rem /hr
n
=
the num ber of half-value layers
0
I/I =
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
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
Dam p soil
8.8
14.3
18.2
20.8
exam ple:
How m any half-value layers of polyethylene are needed
to attenuate a 100 m Rem /hr 5 MeV neutron source to 5
m Rem /hr? How thick does the polyehylene need to be?
0
I
=
I x 0.5
n
I
= 5 m Rem /hr
0
I
=
100 m Rem /hr
n
=
the num ber of half-value layers
0
I/I =
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
