CHAPTER V OF SECRET SOVIET MANUAL ON ATOMIC WEAPONS AND ANTIATOME PROTECTION

• Document Type: CREST
• Collection: General CIA Records
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• Document Page Count: 44
• Document Creation Date: December 22, 2016
• Document Release Date: January 12, 2012
• Sequence Number: 1
• Publication Date: August 16, 1962
• Content Type: MEMO

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2 SE~RET
CENTRAL INTELLIGENCE AGENCY
WASHINGTON 25, D. C.
IRONBARK
1 6 AUG 1962
MEMORANDUM FOR: The Director of Central Intelligence
SUBJECT : Chapter V of SECRET Soviet Manual on,Atomic
Weapons and Antiatomic Protection
1. Enclosed is a verbatim translation of Chapter V of a Soviet
SECRET document entitled "A Guide to the Combat Characteristics of
Atomic Weapons and to the Means of Antiatomic Protection". It was
published in 1957 by the Ministry of Defense, USSR.
2. For convenience of reference by USIB agencies, the
codeword IRONBARK has been assigned to this series of TOP SECRET
CSDB reports containing documentary Soviet material. The word
IRONBARK is classified CONFIDENTIAL and is to be used only among
persons authorized to read and handle this material.
3. In the interests of protecting our source, IRONBARK
material should be handled on a need-to-know basis within your
office. Requests for extra copies of this report or for utili-
zation of any part of this document in any other form should be
addressed to the originating office.
Richard Helms
Deputy Director (Plans)
CSDB-3/650,077
i S RET
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ET
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Original: The Director of Central Intelligence
cc: The Director of Intelligence and Research,
Department of State
The Director, Defense Intelligence Agency
The Director for Intelligence,
The Joint Staff
The Assistant Chief of Staff for Intelligence,
Department of the Army
The Director of Naval Intelligence
Department of the Navy,
The Assistant Chief of Staff, Intelligence
U. S. Air Force
The Director, National Security Agency
Director, Division of Intelligence
Atomic Energy Commission
Chairman, Guided Missiles and Astronautics
Intelligence Committee
Deputy Director for Research
Deputy Director for Intelligence
Assistant Director for National Estimates
Assistant Director for Current Intelligence
Assistant Director for Research and Reports
Assistant Director for Scientific Intelligence
Director, National Photographic Interpretation Center
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Copy No.
40 S ET
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~F'1CRFT
IRONBARK
CSDB-3/650,077
COUNTRY : USSR
SUBJECT Soviet Manual on Atomic Weapons and Antiatomic
Protection (Chapter V)
DATE OF INFO : 1957
APPRAISAL OF
CONTENT . Documentary
SOURCE . A reliable source (B).
Following is a verbatim translation of Chapter V of a
Soviet SECRET document titled "A Guide to the Combat Charac.
teristics of Atomic Weapons and to the Means of Antiatomic
Protection". This manual was published in 1957 by the USSR
Ministry of Defense as a replacement for a similar 1954 manual
(CSDB-35586 ), and is referenced in the Information Collection
of the Artillery (cf. CSDB-3/649,649). It had not been super-
seded as of a e 1961. A similar, more general document was
also published by the 6th Directorate of the Ministry of
Defense in 1969 (CSDB-3/649,686)..
Copy No.
-SA F(V1FT
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ow SECRET
IRONBARK
CSDB-3/650,077
Chapter V
The Penetrating Radiation of an Atomic Burst
Penetrating radiation is a destructive factor which is
peculiar to an atomic burst. It consists of a flux of gamma
rays or neutrons emitted by the burst of an atomic or thermo-
nuclear weapon,
19, General Description and Parameters of Gamma Radiation
Gamma radiation is characterized by:
-the energy of gamma quanta, E , measured as a rule
in millions of electron volts; depending on the energy of the
gamma quanta we can distinguish hard gamma radiation (Elf > 1s MEV)
and soft gamma radiation (E1< 1 NMVT; this distinction is,to
a certain degree, relative;
-a flux of gamma quanta Nr, i . e . , the number of gamma
quanta passing through 1 cm2 of a surface perpendicular to
the axis of propagation of the gamma quanta in a unit of time
(sometimes for the entire duration of the radiation);
-the radiation intensity Iy, i.e., the quantity of
energy borne by the flux of gamma quanta N ; for monochro-
matic gamma radiation the radiation intensity If _
E y x NI MEV/cm2/sec
The intensity of gamma radiation at any distance from
the source of the radiation depends on:
-the activity of the source, i.e., the number of gamma
quanta emitted by the source per unit of time (usually per
second);
-the energy of the gamma quanta emitted;
-the distance from the source of the radiation;
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the form and dimensions of the source of the radiation;
the attenuating capacity of the medium between the
source and the given point. '
The destructive effects of gamma radiation are a function
of its ionizing capacity, which depends not only on the flux
of gamma quanta but also on their energy. The ionizing
capacity of gamma radiation is defined by the magnitude of the
radiation dose, D;. The magnitude of the gamma radiation
dose is expressed in roentgens (r).
A roentgen is that dose of gamma radiation which at 00 C.
temperature and standard pressure will generate 2.08 x 109
ion pairs per cm3 of dry air.
Since 33 ev are expended in the generation of one ion.
pair in air, one roentgen corresponds to 6.86 x 1010 ev,or
0.11 ergs of the energy absorbed by 1 cm3 of air.
The dose per unit of time is called the dose rate R j.
The ratio of the dose rate RY (r/sec) to the gamma
quanta flux N, (quanta/cm2sec) ana their energy Et (MEV) is
expressed in the following formula;
R~ - 1.46 x 10-5 N,E, r/sec. (141)
In this formula, is the linear coefficient of
absorption of gamma radirtion, i.e., the energy fraction
lost by a gamma quantum by ionization along a 1 cm path.
Since it is customary to take air as the medium, the degree
of ionization of which serves as a measure of the radiation
dose or dose rate,_ in formula (141) is the absorption
coefficient for air. Values of for air with various
E, are given in Table 63 and in Fig. 88.
XOs
- 4"_'}---! 50
0 O,il L{u4 QOo c~fjd UIO 0,12 QJ4 fyMPe
e i.e z4 z,z 2,a ~,e ze.:.eA:~x.-1?%~
Fig. 88 Dependence of th
e linear
coefficient of
for air on gamma
absorption ,,,
quanta
energy.
-
,
TS#182471
F
-3-
44~ SF - ET
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IRONBARK
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Values of Linear Coefficient of Absorption fr't' for Air
in MEN
in cm-1X105
inLra V
in cm-1X105
in MMEV
in cm-1X105
0.02
73.4
0.4
3.9
1.8
2.9
0.04
12.0
0.6
3.8
2.0
2.8
0.06
4.5
0.8
3.7
2.5
2.7
0.08
3.3
1.0
3.6
3.0
2.5
0.10
3.1
1.2
3.5
4.5
2.1
0.12
3.0
.1.4
3.2
6.0
1.8
0.20
3.4
1.6
3.1
12.0
1.4
The principal sources of gamma radiation in an atomic
burst are the radioactive fission fragments present in the
zone of the burst, which occupy during the first few seconds
a comparatively small extent, approximately spherical in shape,
and neutron capture reactions by the nuclei of nitrogen atoms
of the air N14 (n ,y ) N15
Gamma rays are emitted even during the process of the
nuclear chain fission reaction. They are, however, to a
great degree attenuated by the massive casing of the atomic
weapon, and they do not play a real role in the overall
gamma ray flux.
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s ec.
Figure 89. Change in the Intensity of Gamma
Radiation with Time for Medium-Yield Weapon Burst
(The broken line shows the change in the
radiation intensity of the fragments).
The intensity.of gamma radiation sharply declines with
time. In Figure 89 the broken line shows the decline of
radiation intensity as a consequence of the rapid decrease
in the overall number of radioactive fragments (mostly short-
lived), occurring as a result of their decay. The solid line
shows the overall drop in the intensity of gamma radiation
of an atomic bomb of medium yield'' occurring as a result of
the decay of the fragments and as a consequence of the rise
of the radioactive cloud, and also as a result of the rapid
(in fractions of a second) decrease in the total number
of neutrons captured by nitrogen nuclei in the air.
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As is evident from the graph, approximately 5 seconds
later, the intensity of gamma radiation Teaching the
earth's surface has dac easood by a factor of hundreds.
Even after ten seconds, however, the intensity of gamma
radiation amounts to tens of roentgens per second.
Therefore it is customary to consider that the time
of action of gamma radiation on surface objects in
medium-yield bursts is about 10 seconds. Where t =
0.5 to 1 sec, a sharp deceleration of tl'e drop in radiation
intensity takes place. This deceleration depehnds.on the
influence of the cavity of rarefied air (rarefied zone of
the shock wave). Gamma rays pass through the rarefied air
cavity almost without attenuation, The higher the yield
of the burst, the greater the dimensions of the rarefied
cavity and the sharper its influence on the ratio
IY = f(t). In high-yield explosions, a strongly pro-
nounced maximum is even observed in the ratio IY= f(t)
corresponding to the time for passage through a given
point of the shock wave compression zone
Ten seconds after a burst the fission fragments of
a single nucleus and the products of their decay emit on
the average 3 to 4 gamma quanta. Hence it follows that in
an atomic bur*t ? with a TNT equivalent of 30 kta during which
about 4 x 1024 nuclei fission, the total quantity of
emitted gamma quanta amounts to
N. = 4 x 1024 x (3 4)X1.5 x 1025 gamma quanta.
The average energy of gamma quanta emitted by.fission
fragments is about 2 MEV. Therefore the ener carried off
by gamma radiation is equal to E- 2 x 1.5 x 1g5 d 3 x 1025MEV 1.1
x 1012ca1, i.e., it consists of about 4 percent of all the
energy 'liberated by the burst.
The average gamma quanta energy emitted in the
N14(n,r)N15 reaction is..equal to approximately 4 MEV,
but their number is approximately equal to the gamma
quanta emitted by fission fragments. However, the
SE T
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"capture" gamma radiation lasts only for tenths of_a
second. During this time the dimensions of the
rarefied air cavity are rather small, and therefore its
influence,on the propagation of.gamma radiation arising
out of neutron capture by the nuclei of nitrogen atoms
is negligible. On the other hand, the cavity exerts a
substantial Influence on the propagation of the gamma.
radiation of fragments in direct proportion to the -
yield of the burst. Fence it follows that the relationship
between the doses caused by "capture" and radiation frag-
ments depend on the yield of the burst: the smaller the
yield of the burst the greater the proportion in the
overall dose of "capture" radiation. In addition, since
the energy of the gamma quanta of "capture" radiation
is significantly higher than the gamma quanta emitted by
fragments, the relationship between doses depends also'
on the distance from the center of the burst: with
increasing distance the ratio of "capture" radiation
grows, since it is more penetrating. At distances in
excess of 1800 to 2000 m almost the entire dose of gamma
radiation is short-term "capture" radiation.
The dose of gamma radiation D at various distances
Pc from the center of an atomic burst can be calculated from
the formula
D y = $2 e-R/250 r,
(142)
where k is a coefficient which depends on the TNT equivalent
OIWSOSRET
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of the bomb; and R/250 is a multiplier which takes into account
the attenuation of gamma radiation by air as a result of inter-
action of the gamma quanta with the atoms of the air.
The coefficient, k, is linked to the TNT equivalent, q
(in kilotons), by the empirical relationship
k = 1.4 X 109q/I + 0.2 (aq)0?65 7 (143)
where a = 2 is the coefficient for a surface burst, and a = 1
is the coefficient for an air burst.
Such a dependence of k on q can be explained in the first
place by the change in the number of fragments proportional
to the TNT equivalent, and consequently of their overall
activity, and secondly by the effect of the cavity of rarefied
air on the propagation of gamma rays. A graph of the relation-
ship k = f (q) is given in Figure 90.
C C~"if CT
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Table 64
Values of the Doses of Gamma Radiation at Various
I'll
s ate paces;
Distance, Dose of gamma radiation, D , in roentgens, for
in meters Lbursts _oaF__f o1iow T1~' eguiv Lents a
200
-300000
-240000
300
83000
66000
400
31500
25000
200000
150000
500
13000
10400
83000
62000
-1000
000
600
6300
5000
40000
30000
-500
000
700
2900
2300
1.8000
13500
200
00C
800
1600
1300
10000
7500
120
000
900
900
700
5500
4100
66
000
1000
460
3 70
3000
2250
36
000
1100
250
200
1600
1200
19
000
1200
140
110
900
680
10
800
1300
80
65
500
380
6
000
1400
45
35
300
220
3
600
1500
180
140
2
200
1600
100
75
1
200
1700
60
45
700
1800
450
1900
2 60
2000
170
2100
100
2200
60
-700000
-340000
135000
82000
45000
25000
13000
7400
4100
2500
1500
820
480
3 40
180
120
70
40
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CSDB-3/650,077
.
4 6 8 -0 - 41 6180100 210 410610IWO 2000 4000
Kilotons
50X1-HUM
Fig. 90. Relationship of coefficient K to TNT equivalent q.
2 .~._
I0 - - 11 : -F - - - - - t
`'
:
-
~-
S
ur
fa
c
e b
ur
st
Air
bur
s
t
--
-
q
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IRONBARK
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-
0
0
V:
r
t]"
00,
C
01
00
0 00
00
00
Figure 91. Dependence of Gamma Radiation Dose on Distance
from Center of Burst
4M SECT
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Calculations, from formula (143) , of the gamma
radiation doses at various distances from the center of
low- (q = 8 kt) , medium- ((I = 30 kt) and high- (q = 150kt)
yield atomic bursts are given in Table 64. A graph of
the relationship Dy = f (It) for surface bursts of the above
weapons is shown in Fig. 91.
The magnitudes given in Table 64 are total doses, i.e.,
doses for the entire radiation time; consequently they are
defined as ` R y (t) dt.
The function R (t), representing the change in dose
rate with time, depeids on the TNT equivalent and the
distance from the center of the burst.
The greater the TNT equivalent the slower the dose rate
decreases with time-and the longer the action of the gamma
radiation. Fig. 92 gives the approximate curves for the
change in gamma radiation dose with time as a percentage of
the total dose for low-, medium- and high-yield weapons.
aQ
IOU
90
0
70
'G0
50
C 40
30
aA 1(i!jv
0
A
Fig. 92. Dependence of gamma radiation dose on time
for weapons:
1.-- of low yield; 2 -- of medium yield;
3 -- of high yield.
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20. General Description and Parameters
o a Neu ron Flux
In an atomic burst neutrons,are generated in the chain
reaction process and during the decay of some fission
products.
The number of free neutrons, not involved in the
chain reaction amounts to 1.5 per fission for uranium-235
and 2 per fission for plutonium-239.
Fission neutrons are emitted several microseconds after
the beginning of the chain reaction and are called prompt
neutrons. The emission of neutrons by fragments continues
for several seconds after the burst. Such neutrons are-therefore
called delayed neutrons.
The greater part of the prompt neutrons, having an
energy of on the average about 1 MEV, are slowed down to
a very low energy level by the casing surrounding the
atomic charge . The maximum energy of these neutrons,
during vaporization of the casing, amounts to approximately
5 KEV. Such neutrons are not propagated over great distances.
Therefore, near the center of the burst '.in a zone with a radius
of 300 to 500 meters, there is formed a "cloud" of neutrons
of great concentration.
.~- SEZ` FT
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The remaining free initial neutrons pass through the
casing without appreciable loss of energy. These neutrons,
just like the residual ones (having an energy of 0.4 to 0.6
MEV), are propagated at great distances from the center of
the burst. Interacting with the nuclei of atoms of air,
they slow down, as a result of which at a given distance
from the center of a burst one can find neutrons of various
energies, down to the thermal level (0.025 EV).
Although the number of residual neutrons emitted from
the zone of a reaction is small in comparison to the number
of initial neutrons (on the average about 0.3 per cent), in
an atomic burst this correlation is greatly altered. Initial
neutrons are attenuated to a considerable extent by the
casing, while residual neutrons are emitted after evaporation
of the casing. In addition to this the propagation of resid-
ual neutrons is facilitated by the formation of the cavity of
rarefied air. As a consequence, the proportion of residual
neutrons in the overall neutron flux of an atomic burst
grows significantly. It has been established experimentally
that they constitute about 20 per cent of a low-yield burst,
about 40 percent of a medium-yield burst and up to 90 per cent
of a high-yield burst.
The spectrum of the neutrons of an atomic burst is
usually divided into three groups: fast neutrons (En > 1 MEV),
intermediate neutrons (100 EV 0.1MEV . The magnitude of a flux
of slow neutrons has the practical value that from it we can
calculate the induced radioactivity of soil, types of
weapons, equipment and other objects.
The total dose of neutrons with energies E > 0.1 NEV
at various distances R(m) from the center of a burst is
defined by the formula
Df = L e-R/250 bray (144)
R2
where m is a coefficient depending on the TNT equivalent;
a graph showing m = f (q) is given in Fig. 93.
*-To- describe the ionizing effects of neutrons, the physical
roentgen equivalent (fre) is used. One fre is that dose of
neutrons the impingement of which on 1 cm3eof material absorbs
95 ergs of energy. 50X1-HUM
-15-
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Approximate values for neutron doses at various distances
from the center of burst of atomic weapons with TNT equi-
valents of 8, 30 and 150 kt are given in Table 65. A graph
of D. = #(R), for the above weapons, is given in Fig. 94.
A flux of slow neutrons can be calculated from the
rough formula
-R/140 2 (145)
Pm n e neutrons/cm ,
where n is a coefficient depending on the TNT equivalent;
a graph of the relationship n = f(q) is given in Figure 95.
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iii
~f7
Fig. 93. Dependence of coefficient m on TNT equivalent q
Ow_SE~BET_
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Q
R
b
~e
t~
00
1
0
ti
_
00
0
Of`
T
r
00
0
-
0
00
00,
7
N
i q
G
10 Qb b tT N a0 to Q Qb to tl N
926
5OX
Figure 94. Dependence of Neutron Dose on Distance from
Center of Burst.
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Table 65
Values of Neutron Doses at Various Distances from the
Center o an Atomic Burst
in meters
Distance
Neutron dose Dn, in bre, for bursts of
the following TNT equivalents
,
8 kt
30 kt
150 kt
100000
33000
165000
12500
62500
5200
26000
130000
2500
12500
62500
1150
5750
29000
800
3150
16000
900
1750
8750
4600
1000
2500
1100
1400
1200
1300
1400
1500
1600
1700
1800
1900
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?
4
lot
n
it)
lo!
2
IC 2.T 10l'
:021SD!?
I
2
10 20 1016
10'0 ,0a
6
4
l o's to
10 n to
_
w 1
!
-
1
i
1. Ili
} + t+
n 1 n'
11;
?-, 44
i I}I
Ile
i {I
1
z
3 b
67
8 20 411 60AU1W 21u 4U16711IS19*W 2 !)Mt 4 I CI M
Figure 95. Dependence of the Coefficients n and n' on
TNT equivalent q.
SECRET
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The formula given for fluxes of slow neutrons is valid
for distances in excess of 400 to 500 meters from the center
of a burst. In the area up to 400 to 500 meters a flux of
slow neutrons changes with distance according to the
following correlation:
Pm ,;, n' e- W50neutrons/cm2. (146)
The values of the coefficient n' for various values
of q are shown in Figure 95. The latter formula for the
change of P in this area can be explained by the fact
that in a given instance a flux of slow neutrons is made
up of neutrons which have been slowed down by the casing of
the bomb, and which create a "cloud" with a great concen-
tration of slow neutrons. At great distances from the
center of a burst, slow neutrons are formed by the slowing
down by the air of fast and intermediate neutrons.
Approximate values for fluxes of slow neutrons, cal-
culated by the formulas given above, are shown in Table 66.
Table 66
Values for Fluxes of Slow Neutrons at Various Distances
from the enter of an Atomic Burst
Distance,
in meters
Flux of slow neutrons (neutrons/cm2) of
bursts of the following TNT equivalents
1
150 kt
100
3.3x1016
2.2x1017
2.2x1018
200
4.6x1015
30x1016
.
30x1017
.
300
6.2x1014
.0x 1015
4
.0x1016
4
400
8.4x1013
5.4x1014
5.4x1015
500
1.1x1013
7.0x1013
7.0x1014
600
1.8x1012
1.3x1013
1.5x1014
700
6.7x 1011
5.3x1012
1.0x1014
800
3.4x1011
2.7x1012
5.1x 1013
900
1.7x1011
1.3x1012
2.5x1013
1000
8.2x1010
6.5x1011
1.2x1013
1100
4.1x1010
3.3x1011
6.1x 1012
1200
1.8x1010
1.5x1011
2.7x1012
1300
9.1x109
7.3x1010
1.4x1012
1400
45x109
.
3.6x1010
6.8x1011
1500
.2xl09
2
1.7x1010
3.3x1011
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IRONBARK
CSDB-3/650,077
21. Attenuation of Penetrating Radiation by Protective
Thicknesses
Attenuation of Gamma Radiation
Gamma radiation interacts with the atoms of the medium
through which it passes, as a result of which the intensity
of the radiation diminishes.
The main types of gamma radiation interaction with the
substance of a medium are: photoelectric absorption,
scattering and the formation of electron-positron pairs.
In photoelectric absorption a gamma quantum gives up
its entire energy o t e e ectron of an atom. A part of
this energy (generally 30 to 50 EV) is expended on knock-
ing the electron out of the electron shell of the atom, and
the remainder is transformed into kinetic energy of the
electron. Photoelectric absorption is defined by the linear
coefficient, Mfe cm- , which is also often expressed as
With scattering as a result of a collision with electrons,
gamma quanta `give up to the electrons a part of their energy
and change the direction of their motion, i.e., they are
scattered. Scattering is expressed by the coefficient Mras(cm-1),
with the equation _
lu - /4 r/ /up
where / is a coefficient defining the amount of energy of
gamma radiation which is carried a distance of 1 cm by the
scattered gamma quanta; and MJ2 is a coefficient defining the
amount of energy of gamma radiation which is absorbed within
1 cm during the scattering of gamma quanta, i.e., the energy
transmitted to the electrons of atoms.
ON11111h. ~r
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IRONBARK
CSDB-3/650,077
~
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0
p
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S1S PFT
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- SF(SQFT
IRONBARK
CSDB-3/650,077
The formation of pairs is also a process by which a gamma
quantum is transformed into' two particles -- an electron and
a positron. The formation of pairs occurs only when the energy
of gamma quanta exceeds 1.02 MEV. The attenuation of gamma
radiation as a result of the formation of pairs is defined by
the coefficient,/ (cm-1).
The complete linear coefficient of attenuation of gamma
radiation,/{, equals
P = jLfe- /bras / /r ar
The inverse magnitude ofj4 is known as the average free
travel of gamma quanta, X (cm) .
The values of /k and T for several substances and for
various energies of gamma quanta are shown in Table 67.
For those materials not shown in Table 67, is defined
by the correlation
~x - ? P (147)
where /I is the linear coefficient of attenuation of a substance
with a densityf in g/cm3 (determined from Table 68); and
is the density of the substance for which/4 is being determined.
A more exact correlation can be obtained if, instead of
density in g/cm, we take the electron density, i.e., the number
of electrons in 1 cm3 of a given substance. Electron density,
P-, , can be determined from the formula
R.=b02x/O''3 O. t (148)
where a.i is the relative proportion of a given element in
the substance (by weight);
2j is the charge of the nucleus of a given element;
Ai is the atomic weight of a given element;
P is the density of the substance in g/cm3.
The electron densities of several substances are given
in Table 68. 50X1-HUM
-24-
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IRONBARK
CSDB-3/650,077
Table 68
Electron Densities of Certain Materials
Material
ens i
y
in g/cmi
Electron ens ,
in electrons/cm
Air
Wood
Water
Earth
Concrete
Aluminum
Iron
0.00129
0.7
1.0
3.89x1050
2.24x1023
3.34x1023
4.85x1023
7.30x1023
7.86x1023
2.2x1024
Attenuation of a dose of gamma radiation with a narrow
monochromatic flux follows the exponential equation:
D = Do a-Ph ,
(149)
where D is the gamma radiation dose after passage through
a medium of h thickness, in centimeters; and D.o is the gamma
radiation dose in front of the medium.
Attenuation of a broad monochromatic flux of gamma rays
follows the more complex equation:
D = Do e-/thVh, (150)
where Vh is a coefficient taking accourtcf the increase of
the dose w it h he thickness of the material h(cm) as a result of
the action of radiation scattered within the thickness.
For gamma radiation with an energy of about 1.8 to 2
MEV, the magnitude of Vb. is determined from the following
formula:
-25-
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IRONBARK
CSDB-3/650,077
for media consisting of the atoms of light elements
(air, wood, earth, concrete, brick, etc.),
Vh = 1 ? 0.44/ Oh / 0.015/00
; (151)
for iron (armor)
Vh - 1 ,e 0.15 h / 0.002 h2; (152)
for lead
Yh = 1 , 0.1h,Z 0.002 h2
(153)
The gamma radiation from an atomic burst is not homo-
geneous: it consists of gamma quanta of various energies,
so that with the increase of distance an ever-growing pro-
portion of the overall gamma radiation flux is taken up by
scattered gamma radiation, which has a lower energy and con-
sequently a lower penetrating capability. For this reason
the exponential equation for the diminution of the dose is
valid only beginning from a certain thickness of material,
h . At the surface a layer of material with a thickness
of ho undergoes a sharper f a l l o f f in the dose as a result
of the attenuation of the softer scattered gamma radiation.
Beginning from ho the rule for the change in the dose
of gamma radiation can be determined from the equation
D = aDoe-PSh 9
(154)
where a is a coefficient taking account of the attenuation
of the softer scattered radiation in the thickness ho; and
/Vef is the effective coefficient of attenuation.
Since the spectrum of gamma radiation changes with
distance, it becomes softer with an increase of distance with-
in defined limits, so that even the magnitude, is also a
function of the distance from the center of a Uuu'rst. The
graph in Figure 96 shows the relationship uef = f (R) for earth.
Using the correlation (147), oefcan also be determined for
other materials.
-26-
Cr-k'Dr-T
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IRONBARK
CSDB-3/650,077
600 ?D0 8lm 0001000 uoG
Figure P6. Dependence oY efon Distance (for earth)
T- - ~- T-4
Wood
Earth
Concrete
1 Iron
zUt3 a IOU X0(1 GUt)'Iikl6w UUU Iwo 1w.l,'.UU~JuVt+CJlyxl(
in meters
Figure 97. Dependence of dpol on Distance from the Center.
-27-
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IRONBARK
CSDB-3/650,077
Formula (154) can be represented in another way by
using the half thickness, that is the layer which atten-
uates a dose of gammaradiation by half.
Half thickness equals
d _ 14~93 (155)
Formula (154) then takes the form
D = aDo2-h/dLC. .
(154)
Graphs of the dependence of d for earth, concrete,
wood and steel and of the coefficient. a, on distance from
the center of a burst, H, are given in Figures 97 and 98.
,0. 1
/.0
0.6
0,4
-rr-+H-3;
0
2X) 3, 44 Soo GOO 7D J'O?od io~orto~ g,e,. w' oo'R, in meters
Figure 98. Dependence of Coefficient a, on Distance from
Center (or Ground Zero) of a Burst
In Figures 99 and 100 graphs are given of the dependence
of the coefficient of attenuationbf gamma radiation
Do aAtfh
K = = a
for earth, concrete, wood and steel for minimum and maximum
values of d-2 .
>2$o
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IRONBARK
101
?
to
s
Figure 99. Attenuation of Gamma Radiation by Concrete,
Earth and Wood
_29_
dftb SFNtFT
ti
.
~M~
-a:~~ - ~,
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AMLAF~Kpr-T
IRONBARK
CSDB-3/650,077
n R_ : ?3 2c ?2om
Figure 100. Attenuation of Gamma Radiation by Iron
-30-
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SEC2FT
IRONBARK
The attenuation of a neutron flux which occurs throw h t
Attenuation of a Neutron Flux
I &4= u
interaction with atomic nuclei, is defined by the cross-section of the
nuclear reaction c; expressed numerically as the probability of the
interaction of a neutron with a nucleus.
There are two basic forms of interaction: scattering of neutrons
and their capture by atomic nuclei.
Scattering is either elastic, like the collision of two spheres,
or inelastic, in which case the neutron penetrates the nucleus. A
nucleus thus excited passes in a very short interval of time to a steady
state, emitting a neutron of lower energy and a gamma quantum.
Scattering is the typical form of interaction for fast and intermediate
neutrons. Elastic scattering can be observed with the interaction
of these neutrons with any nuclei, and inelastic scattering with
the heavy nuclei.
The capture of a neutron by a nucleus leads to a nuclear reaction.
This form of interaction is typical for slow neutrons.
Depending on the type of interaction, we have the following cross-
sections: for elastic scattering d ur ; for inelastic scattering dciur;
and for capture G' zakh
Cross-sections of interaction 6 ur, far and aZ , are generally
measured in cm2 or barns (one barn ? l T'-24c'i).
Just as the destructive effects of neutrons emitted by an atomic
burst depend on fast and intermediate neutrons with an energy of
En > 0.1 NEV, we likewise consider only the scattering processes in
calculating the attenuation of a neutron flux.
Shown below are approximate methods for reckoning the attenuation
of a neutron flux by various materials.
Attenuation of a Neutron Flux by Earth, Wood, Concrete,
Brick and Other I terials
Elastic scattering of neutrons is typical of these materials, since
they consist of atoms of the light elements. In an elastic collision a
neutron transfers part of its energy to a nucleus, and changes the
direction of its motion.
The change of energy of a neutron is defined by the coefficient
which equals
E
In El (156)
where Eo and El correspond to the initial and final (after collision)
energies of the neutror, respectively.
-3
q0b"SFT
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IRONBARK
CSDB-3/650,077
In agreement with the rule for the con i=i.n of -last
c
-1 -f- p-z2? (157)
where A is the atomic weight (mass number) of the nucleus.
The average angle of scatter can be determined from the
correlation
cos 8 = 2 (158)
If the material consists of several elements, then
= Nip and (159)
cos e = E N. cos 61 ,,
n Ni
In these formulas Ni is the number of nuclei of a given
element per cm of the material, and is determined thus
_ 6, a2 x /O Qi P (161)
L'
where AL is the atomic weight of a given element;
P is the specific gravity of the medium; and
'Li is the percentage of a given element in the medium.
Considering that neutrons emitted by an atomic burst have
an energy approximating 2.5 MEV or lowerf* and that they will
have destructive effects so long as their energy does not
drop below 0.1 MEV, we can find the path along which neutrons
with an energy of E6=--:--2.5 KIEV will be slowed down to an energy
of El = 0.1 AEV as a result of elastic collisions with the
atoms of the medium. This path, which is called the retardation
length L, can be approximately determined from the formula
L _ r (162)
or elements where to an accuracy of 1 percent
For example, for iron (A56)E0.0357.
** There are high-energy neutrons, but their number is quite
small.
-32
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{RONBARK
CSDB-3/650,077
3 ( cos-6 X u)(163)
rk4
4Q
Here u -Z = 3, 2-
E and consequently c ul = 1n~? = 1-n 0 'S
The function entered under the integral sign represents
the square of the mein free iit of a neutron
from one elastic collision to another.
For a neutron of a given energy the mean free
?`r a.5 1 (164)
Ner v~
For a compound of elements""
Xr~as= 5'N,o-,
U'i-
approximate calculation fom formula (163) we can
simplify by treatingAras as a constant. 3~ CI _ .2-
3, =
-cos ) 3~~_COs9
L _ ~- ~oS e
~~ bras
Attenuation of a flux of fast neutrons, and consequently
the diminution of their dose, can be determined from the
formulas:
/a_ Pa-411- _ p -~ldP_/ (167)
='O0 (168)
where h is the thickness of the material.
Hence the coefficient of attenuation will equal
(169)
-33-
. cF\evF,
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SE
IRONBARK
CSDB-3/650,077
The principal magnitudes in the formulas given above,
which are required for the calculation of the attenuation
of fast neutrons by various elements, are shown in Table 69.
Basic Data on the Retardation Capacity of
Various Elements
Atomic
Weight
i-cos 0
Hydrogen
1.00
0,333
Carbon
0.15
8
0.944
Oxygen
0.12
0
0.958
Sodium
0.08
5
0.970
Aluminum
0.073
0.975
Silicon
0.07
0
0.976
Potassium
0.05
0
0.983
Calcium
0.049
0.983
Average values
for Our , in barns
ranging from 0.1
to 2.5 1dEV
Values for retardation lengths L and for half value layer
d/,a/, calculated from the above formulas for several material
are given in Table 70.
Table 70
Values for Retardation Lengths L and Half Value Layer dam/
for Several Materials
Water
Wood
Earth
Concrete
Density
P
in g/cma
1.0
0.7
1.7
2.3
Rtaraton
length L,
in cm
4.5
14
17
12
FIC'RFT
Half 'aTue
Layer dip. o /
in cm
31 ';50X1-HUM
9.I
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IRONBARK
CSDB-3/650, 077
Values for L and dpol , calculated for materials of
particular chemical composition, are shown in Table 71.
Table 71
Chemical Composition of Several Materials
(principal a ements
Material
LH
C 0
Na AlI Si
K Ca
Water
11.1
--
88.9
-- -- --
-- --
Wood
6.3
49.5
44.2
-- -- --
-- --
Earth
1.1
1.4
52.9
0.8 6.5 31.4
2.2 0.4
Concrete
1.1
--
49.2
-- 2.0 26.7
I -- 21.0
light materials.
Attenuation of a Neutron Flux by Heavy Materials
To determine the attenuation of a neutron flux by heavy
materials one must take account of scatter, both elastic and
inelastic. The loss of energy by a neutron is considerably. greater
from an inelastic than from an elastic collision. As a' result
of an inelastic impact a'fast neutron will lose on the average
about 90 percent of its energy, while with an elastic impact,
e.g., with the nucleus of an atom of iron, it will lose only
about 3.5 percent.
The most important practical application is to thick-
nesses of armor plating. Armor plate is composed of 90 to
95 per cent iron, Fe56. Therefore, in calculating the atten-
uation of a flux of fast neutrons by armor plate, one must
assume that the principal role in the attenuation of the fast
neutrons is played by the process of scattering by the nuclei
of iron atoms.
The retardation length for elastic scatter
is determined by the same method as was described--above for-
-35-
4M SE ET
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IRONBARI<
CSDB-650,077
In the range of energies from 2.5 down21o $.1 MEV, bur
is roughly constant and is equal to 3 x 10' cm (dur Q barnsl.
The number of nuclei in 1 cm of iron equals
N = 6.02 x 1023 p = 6.02 x 1023 x 7.8 - 8.4 x 1022 nuclei/cm3.
A 56
x
Nc7 .
and the magnitude
1 4 cm
8.4 x 1022 x 3 x 10_24
1.06 x 42 490cm2.
(1-cos0) 0.0357(1-0.012)
The retardation length for elastic scatter is
Lur = 'V--r = 22 cm.
The cross-section of inelastic scatter for iron
equals o- = 1.16 x 10-24cm2. The average length of
free travel between two inelastic collisions is
Lnur.
No'nur = 8.4 x 1022 x 1.16 x 10-24
The attenuation length for fast neutrons, taking
account of elastic and inelastic scatter equals
L = Li Lnur = 6.9 cm,
Lur+Lnur
and the half value layer is
dpol = 0.693L = 4.7 cm.
Thus the attenuation of a dose of neutrons by armor
plate can be determined roughly by the formula
D = Doe-h/6.9 = Dot-h/4`7, (170)
where h is the thickness of the armor plate, in cm. 50X1-HUM
4W SE ET
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T
IRONBARK
CSDB-3/650,077
By an analogous method the attenuation l ngth for
lead (v" ur .~ 5 x 10-24 cmZ , d' nom. = 1.7 x 10-2' cm2) equals
L,--12.5-cm, and the halfvalueTlayer dp~1 - 8.7 cm. The
attenuation of a dose of neutrons by lead can be estimated
by the formula
D=Doe-h/12.5= Do2-h/8.7. (171)
22. Scattering of Penetrating Radiation in Air
During their propagation, gamma rays and neutrons interact
with air and change the direction of their motion. The process
result of
of radiation scattering repeatedly, as a
it acts on an irradiated object
the burst, but from all other directions as well.
Figure 101. Dependence of nj and nn on angle ce
DoY2
Figure 101 shows the dependence of n r DoY and
nn DID on angle c~>, which is formed by the axis of the
solid angle and the direction to the center of the burst.
In these equations Do and Don are the doses of gamma
radiation and neutron on open terrain at a given distance
from the center of the burst; DOY2 and D n2 are the doses
of gamma radiation and neutrons produced ~y the action of
radiation occurring at a given point from a solid angle equal
to one steradian
-3.7-
SRET
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IRONBARK
CSDB-3/650,077
The coefficients ny and nn in most cases also depend on the
distance from the center of the burst, since with increase of dis-
tance the proportion of scattered radiation also increases. How-
ever, beginning at a distance of 200 to 250 meters the dependence
of n-r and nn on R becomes insignificant and may therefore be
disregarded.
S
The solid angle t for a given point is defined asf2 - ,
in steradians, where S is the area of a spherical sector limited
by the lines of intersection which form a solid angle with a sphere
of radius R.
A dose of scattered radiation, passing through an aperture,
is characterized by solid angle-(L, estimated from the formula
D = Do+ Don D. = Doyn _Yn+ . (172)
23. Method for Estimating the Protective Properties of
true ures and Equipment
The estimation of the protective properties of structures
and equipment from the effects of penetrating radiation has as
its goal the determination as to whether in an atomic blast the
dose of penetrating radiation inside structures or in places
where equipment is located exceeds the permissible dose. Along
with this one usually also takes account of the situation where
a structure or equipment are lccated at an extremely close
distance to the center (or ground zero) of a burst, which deter-
mines the resistance of the structure or equipment to the impact
of the shock wave.
For a given distance calculated from Tables 64 and 65 the
doses of gamma radiation and neutrons can be determined for open
terrain.
Enclosed Structures. The protective thickness of a structure
will 'at enua e a dose' o gamma radiation by a factor of k-c, and
the dose inside the structure will therefore equal
Dr = Dk
r
and a neutron dose :will equal
,
Dn = Don
(173)
(174)
The values of the coefficients of attenuation ky and kn are
determined by the method butlinedin Para. 21.
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The coefficient of attenuation of penetrating radiation,
i.e., the total dose of gamma radiation and of neutrons, equals
Doa, + Don (175)
kpr x r = D r + Dn
Since a neutron dose Don is on average 30 per cent
of a gamma radiation dose, the formula given may take the
form
~, Do7 + 0.3Dor = 1.3kyk? (176)
kpx x DoZ + 0.3Dc, ' kn - 0.3 k7
k,y kn
In the same way one can estimate the protective properties
of tanks. However, one must keep in mind that for the calcula-
tions,k Y and kn, the protective thickness is equal not to some
specific layer of armor (frontal, side, etc.) but to a certain
effective magnitude, which,on the average (according to experi-
mental data) for IS-3 tanks equals 8.5 cm, for T-54s--8.0 cm,
for T-34's--5.7 cm and for PT-76s--1.0 cm.
Open structures and structures with apertures. For the
estimation o t e protective properties o suc structures the
total flux of penetrating radiation is divided into two parts:
the first part is the gamma rays and neutrons which fall on
the protective thickness and become attenuated by it while
penetrating into the structure; the second is the gamma rays
and neutrons which enter the structure through apertures and
are therefore not attenuated. (Obviously for open structures
such as trenches and connecting trenches one has to take
account only of the second type of penetrating radiation.)
If the first type of penetrating radiation creates at
ground level a gamma radiation dose Do~?s , and a neutron
dose Don2, and the second type, doses of Do a and Don Z, then
the total dose of penetrating radiation will equal
Dopr x r = DOV + Don = (Doi-! + Donl) + (D?r,Z-+-Don;L) (177)
The part of the radiation which enters a structure through
apertures depends on the orientation of the structure relative
to the center of the burst and on the magnitude of the solid
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angle-formed by the contour of the aperture. This part of
the radiation can be determined from the graph'in Figure 101,
which shows the dependence of the correlation
nY = Doy2 and nn - Dpn2 (where the solid angle
?T Do
equals one steradian) on angled, which is formed by the axis
of the solid angle and the direction to the center of the burst.
The magnitude of the dose inside a structure, depending on
the radiation entering through apertures, equals
Doy2 = DoIny-M; (178)
Don2 = Donnnfl, (179)
The magnitude of the dose inside a structure, which is
caused by radiation passing through the protective thickness,
is determined by the method outlined above for enclosed struc-
tures. However, one must subtract from the dose at ground
level the magnitude of the dose entering the structure through
apertures. Consequently
DY t -- - ?Y2 D? ( n)
D Y
Dn1 Don - Don2 Don 1-Hann)
- -1 _ -
n
The total dose of gamma radiation inside a structure will equal
D~, D,`1 -t DOY 2 = [l -i- nyfl(k1 - 1) , (182)
r
and the total dose of neutrons
Dn = Dn1 fi Don2 = Don [i + nnII (kn - 1)) . (183)
kn
Consequently, the total dose of penetrating radiation
-inside a structure will equal
Dprr x r ?Dk [l + nyfl (kY - 1] + Don [1 f nnfl(kn - 1)] . Q.84
Y n
Considering that the dose of neutrons amounts to about 30
percent of the dose of gamma rays, we find that
Dpixr =D011+ nftk)= +0.3 [1 nn (kn - l1j? (185)
k kn
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For rough calculations for the majority of structures,
except those with armor plating, the effect of neutrons can
be disregarded, since their dose is relatively quite small,
and the protective properties of such materials as earth, wood
and concrete against gamma radiation and neutrons are approx-
imately the same. Therefore the coefficient of attenuation
of penetrating radiation for such structures can be approxi-
mately determined from the formula
kpr x r ^ 1+k!z . (186)
24. Penetrating Radiation from the Burst of a
ermonuc ear weapon
The typical reaction on which the blast of a thermonuclear
weapon can be based is the fusion reaction of deuterium with
tritium 1D2 1T3 2He4 nl.
0
Neutrons formed by this reaction have an energy of about
14 MEV. Neutrons with an energy of 14 MEV are often called
"super-fast". The presence of a flux of "super-fast" neutrons
is a peculiarity of the penetrating radiation from the burst
of a thermonuclear weapon.
Besides this type of thermonuclear reaction, it is also
possible to make use of the synthesis reaction of helium from
lithium and deuterium
3Li7 1D2 -----~ 22He4 ?ni .
The energy of neutrons emitted by this reaction is also
about 14 MEV.
One of the varieties of atomic weapons is the hydrogen-
uranium bomb (or other type of w ea,p o n ) , in which, as was
already shown in para. 7, there occurs a fission reaction of
the atomic nuclei of natural uranium (or more precisely, of
uranium-238) by neutrons which have been formed by the thermo-
nuclear reaction.
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the composition of the fission fragments of uranium-238
Si
nce
is approximately the same as for uranium-235 and plutonium-239,
it can be assumed that the main proportion of penetrating radi-
ation from the burst of a hydrogen-uranium bomb consists of
gamma radiation from the fission fragments of uranium-238, and
to a lesser degree of a flux of fast neutrons from the fission.
In the overall flux of penetrating radiation, the role of
super-fast neutrons for a given type of bomb is usually insig-
nificant. Therefore the gamma radiation and neutron dose of
a burst of a hydrogen-uranium bomb can be determined from the
formulas given for an atomic bomb in paras. 19 and 20. As an
example, Table 72 gives the data (calculated from the following
formulas) for gamma radiation and neutron doses for the burst of
a hydrogen-uranium bomb where the TNT equivalent, q, equals
1000 kt:
4 x 1012 e-R/250r (187)
D-t ---
D? 3 x 1012 a-R/250 bre. (188)
= R2
D 4 x 1011 a-R/150 bree (189)
s aN R2
Approximate Values for Gamma Radiation and Neutron
Doses for the Burst--of a om , ere q = 1000 t
Table 72
Distance,
in meters
Gamma radiation
dose, in roentgens
Fast neutron
dose, in bre
Super-fast neutron
dose, in bre
1000
-650,000
60,000
500
1250
170,000
12,000
70
1500
45,000
1,000
7
1750
12,000
300
--
2000
3,300
90
--
2250
1,000
20
--
2500
250
6
--
2700
65
2
The attenuation by various materials of the gamma radiation
and of the fast neutrons of the burst of a hydrogen bomb is 50X1-HUM
determined by the method described in paras. 21 and 23.
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