SOLUTIONS AND ANSWERS TO 1996 ABHP EXAM
QUESTION 14
GIVEN: You are asked to estimate the dose to the tissue of an animal that was injected with a
beta emitting radioisotope of phosphorous having a beta yield of 100%. Additional
information includes the specified maximum energy Emax in MeV, specified average
energy Eavg in MeV, radiological half-life Tr in days, approximate absorption
half-value layer X1/2 in mg cm-2, equation for beta particle range R in mg cm-2,
equation for fraction f of beta energy converted to photons, biological half-life Tb of
19 days in soft tissue, atomic numbers Z, and for air and photons with energy E of
0.7 MeV: the mass attenuation coefficient µ/' of 7.56 m2 kg-1 and mass energy
absorption coefficient µen/' of 2.93 m2 kg-1.
Comment: The given equation shown below for the beta particle range R was
incorrectly reproduced from the expression given in the Radiological Health
Handbook. The constant 0.954 shown in the expression for the exponent of E should
have been 0.0954. The values given for the mass attenuation coefficient µ/' and mass
energy absorption coefficient µen/' are 1,000 times the actual values. These incorrect
data are used in the solutions shown below. Answers calculated from the correct data
and equation for the range are shown after the incorrect answers.
SOLUTIONS AND ANSWERS(*):
A. The dose D in Gy to an organ of a live animal is calculated for an initial concentration C(0)
of 1.35 µCi g-1 (5x107 Bq kg-1) over a period T of 10 days for 32P (Eavg of 0.695 MeV and Tr
of 14.3 days) from the following equation using the values for the symbols and bolded units
shown above and for k:
I assume the given biological half-life Tb of 19 days applies and that the average beta energy
is completely absorbed in the organ.
k total removal rate constant = (ln 2)/Tr + (ln 2)/Tb = 0.0850 day-1.
*
D 0.511
C(0) Eavg
k
( 1 e k T ) 3.23 Gy.
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SOLUTIONS AND ANSWERS TO 1996 ABHP EXAM
B. The shallow doses D to the fingers of a person wearing gloves having a mass density
thickness of 10 mg cm-2 is calculated for either 32P or 33P over a time T of (1/6) h for a person
holding an organ containing either radioisotope with an associated energy spatial equilibrium
beta dose rate D
of 0.5 Gy h-1 to the organ:
I assume in either case 1 or 2 below that radioactive decay can be neglected, that the given
half-value layer X1/2 applies to the attenuation of the respective 32P or 33P beta particles
through a total thickness X of 17 mg cm-2 (gloves + 7 mg cm-2 of skin) provided this
thickness does not exceed the range of beta particles, and that the dose rate at the surface of
the gloves is ½ of the energy spatial equilibrium dose rate D
to the organ. In either case, the
shallow dose D is then calculated by the following equation using the values for the symbols
and bolded units shown here and below:
1 D D T e
2
ln2 X
X1/2
.
1. The shallow dose D for 32P is calculated for the given X1/2 of 150 mg cm-2:
*
1
1
D (0.5 Gy h 1)
h
2
6
e
ln2 (17)
150
0.0385 Gy ,
where the range R calculated from the given equation is 617 mg cm-2, which exceeds the
absorber thickness X of 17 mg-2:
R 412 (E)(1.265 0.954 ln(E)) 412 (1.71)(1.265 0.954 ln(1.71)) 617 mg cm 2 ,
which is calculated as 790 mg cm-2 from the correct equation.
2. The shallow dose D for 33P is calculated for the given X1/2 of 6 mg cm-2:
*
D 0,
because the range R calculated from the given equation is 11.2 mg cm-2, which is less
than the absorber thickness X of 17 mg-2:
R 412 (0.249)(1.265 0.954 ln(0.249)) 11.2 mg cm 2.
For the correct data and range equation, the range R is calculated as 59.0 mg cm-2 and the
33
SOLUTIONS AND ANSWERS TO 1996 ABHP EXAM
dose is calculated:
*
*
1
1
(0.5 Gy h 1)
h
D 2
6
e
ln2 (17)
6
0.00585 Gy ,
C. Attenuation of 32P beta particles is approximately exponential over thicknesses less than the
range because the beta particles are emitted with energies from zero up to the maximum
energy.
D. The bremsstrahlung dose rate D
in air is calculated at a distance d of 0.30 m from a mass m
of 0.050 kg of an organ (Z of 7.1) having a concentration C of 5x107 Bq kg-1 of 32P assuming
that the energy E of each photon is the average beta energy Eavg of 0.7 MeV for which air has
the given incorrect µen/' value of 2.93 m2 kg-1:
The fraction f of beta energy converted to photons is calculated from the given equation:
f 3.5x10 4 Z E 3.5x10 4 (7.1)(0.7) 1.74x10 3.
The dose rate D
in air is then calculated neglecting the very slight photon attenuation in air
and in the tissue mass:
m C E f
D 4 % d2
µ en
'
(1.6x10 13 j MeV 1)
1 Gy
j kg 1
3,600 s
;
h
(0.05 kg)(5x10 7 s 1 kg 1)(0.7 MeV)(1.74x10 3)(2.93 m 2 kg 1)
D 4 % (0.3 m)2
(1.6x10 13 j MeV 1)
*
1 Gy
j
3,600 s
h
kg 1
4.54x10 6 Gy h 1 ,
which is calculated as 4.54x10-9 Gy h-1 from the correct value of 0.00293 m2 kg-1
for µen/'.
34