Physics and Radiobiology of Nuclear Medicine

(16.1)

where Gis the exposure rate constant of the radionuclide.

Problem 16.1

Calculate the radiation exposure at 25 cm from a vial containing 30 mCi
(1.11 GBq) of^201 Tl.

Answer
The exposure rate constant G 20 of^201 Tl is 0.45 R · cm^2 /mCi · hr at 1 cm from
Table 16.2. Therefore, using Eq. (16.1), at 25 cm

Because G 20 of^201 Tl in SI units is 12.16mGy·m^2 /GBq · hr at 1 m,Xfor 1.11
GBq of^201 Tl at 25 cm is

It should be pointed out that because the patient is not a point source, the
exposure rate does not vary exactly as the inverse square of the distance.

Shielding

Various high atomic number (Z) materials that absorb radiations can be
used to provide radiation protection. Because the ranges of a- and b-
particles are short in matter, the containers themselves act as shields for
these radiations.g-Radiations, however, are highly penetrating. Therefore,
highly absorbing material should be used for shielding of g-emitting
sources, although for economic reasons, lead is most commonly used for
this purpose. The half-value layer (HVL) of absorbent material for differ-
ent radiations is an important parameter in radiation protection and is re-
lated to linear attenuation coefficient of the photons in the absorbing
material. This has been discussed in detail in Chapter 6.

Problem 16.2
Calculate the number of HVLs and the amount of lead necessary to reduce
the exposure rate from 100 mCi (3.7 GBq) of^131 I to less than 10 mR/hr at 10
cm from the source. (G=2.17 R · cm^2 /mCi · hr at 1 cm and 1 HVL =3 mm of
lead).

Answer

Exposure at 10 cm mR hr

2170 100

10

= 2 2170

×

=.

X=

×

()

=

111 1216

025

215 96

2

..

.

. mGy hr

X=

×

=

30 0 45

25

2 21 6

.

.mRhr

X

n

d

=

G

2

276 16. Radiation Regulations and Protection