NUCLEONICS DATA SHEET No. 25 
Gamma-Ray Shielding 
Attenuation of y-Rays from an Infinite Plane 
mM 
cm! 
0.2 
O. 
0.08 
0.04 
0.03 
244 
0.7 
in. 
80 
60 
40 
30 
20 
0.8 
04 
Q3 
0.2 
cm 
200 
100 
80 
60 
40 
30 
20 
0.6 
0.5 
By M. G. CHASANOV and 
M. SHATZKES 
Applied Research and Advanced 
Development 
International Business Machines Corp. 
Owego, New York 
FoR AN INFINITE PLANE source of 
gamma rays the attenuation factor 
¢/¢o for materials other than lead is 
given by 
; [e* + E1(b)] (1) 
where £,(0) is a form of the exponential 
integral (1). In the case of lead, the 
factor is 
1 
5 (2) 
[e+ 0) 
2 
This nomogram solves Eq. 1 for ele- 
ments other than lead and for gamma- 
ray energies to 3 Mev. For lead, one 
can readily determine the attenuation 
from Eq. 2. 
To calculate attenuation of gamma 
rays from an infinite plane source, it. 
is necessary to have values for the 
linear absorption coefficient u. The 
linear absorption coefficients can be 
obtained from curves (2) or tables (3) 
of mass absorption coefficients by 
multiplying by the density of the shield 
material. 
For a shield thickness t, a quantity b 
is defined as b = wt. For most ele- 
ments other than lead, the value 1 + b 
for gamma energies up to 3 Mev (2) 
can be used as an approximation for 
the buildup factor B. 
Example 
Problem: How thick must an iron 
shield be to attenuate the gamma rays 
from an infinite plane source by a fac- 
tor of one-tenth? 
Procedure: Employing the linear 
absorption coefficient for iron given by 
Glasstone (3), uw = 0.27 cm™, draw a 
line from yp = 0.27 through $/¢o = 
—0.1 and read on the ¢ scale the value 
6.9 cm. 
BIBLIOGRAPHY 
1, T. Rockwell III, ed., ‘‘Reactor Shielding De- 
sign Manual,”’ pp. 374-9 (McGra v-Hill Book 
Co., Inc., New York, 1956) 
2. D. G. Chappell, nucteonics 14, No. 1, 40, 
(1956) 
. S. Glasstone, ‘Principles of Nuclear Reactor 
Engineering,” p. 606 (D. Van Nostrand Co., 
Inc., New York, 1955) 
& 
