Appendix A
DERIVATION of ALTITUDE ABSORPTION
of GAMMA RADIATION
Keran O’Brien, Radiation Branch, Health and Safety Laboratory
The equation giving the dose rate above a hole in an
infinite half-space that subtends an angle 4°, when the
half-space is uniformly contaminated with a gamma
where
j is the disintegration per second per cubic cen-
tumeter and E, is the average source energy.
1 v = of
om, ‘ ay
20Y
emitter, is described in Reference 10 and is:
=
y
Where:
E
20Y
0
i
A.
Ath, 8°)
(A-1)
E is the gamma energy emitted per cubic centimeter by the contaminant
(A.8)
.
The constants may be converted to appropriate units
to relate contamination density to gamma dose rate by:
Ge (3,600)
K = ————__
EQ
Wee
ois the density of the absorbing medium
(A.9}
his the height of the detector, in meters, and
Y= ft the ratio of the total attenuation coBe
Where:
efficient to the energy absorption coefficient
4.8 x 10 Mesy
He = 3.54% 10 %cm"! (for water}
of the medium, corresponding to the source
W = 3.25 = 107° Mev ( (32 5 ev )
energy
3,600 sec/hr
For A:
8
c = 3.7.x 10" (photons/sec)/m?
1
.
A(b, 6°) = ; {tuEi (—tu)+e
—tu
1
B(ta)}
Then:
The dose rate above a plane, similarly contaminated,
can be obtained by the partial derivative of Equation A.1
to obtain an infinitesimal thickness of slab:
at
oh dh = Ip
=
v
where Cy
with M(tu) = —E, (—tu) + eT B(tu)—B! (w)—1])
Where:
B'
(A.4)
(A.5)
= Sa
i
For the case of radiation from water or land contaminated with fission products, seen byan aircraftmounted detector, a finite diameter of contamination
on the surface is described vy @ hali-angle sensitivity,
6.
CASE I.
Water contamination from Equation A. l.
Lh, 6) = Ah, @°)-Ath, 6)
(A.6)
and
EL
Eg]
Boy * Boy
'
Cc Lh, 6°)R/hr
v
{A.10)
Land Contamination:
Jth, 6) = Mth, 6°)-—Mt¢h, 8)
(A.13)
HyEdh =
(A.12)
and
20Y
HeEoK
20
where k represents disintegrations per second per
The clearing on the surface also subtends on angle @.
4
‘
j
289
2
= curies per cubic meter.
CASE Il.
oY
0.3549
(A.3)
This is:
Ip = oh dh M(tu)
°
108 cm3/m!, and
E, is assumed to be 0.5 Mev
(A.2)
t = pyh, u = sec 6°, and Bitu) is a polynomial
-
(A.7)
square centimeter.
This reduces Equation 4.4 to:
HeE gk d(h, 8")
0
ly ~ “5
(A.13)
With the constants converted to appropriate units as in
Case J, and 104 cm?/m’.
¢ GHe(3,600k
3,600)k
K = er
Ep
DCE ANCHIVES
Ip = 3.4427, J(h, éR/hr
(A.14)
(A.15}
where Cp = curies per square meter.
—
ot