99 Mo–99mTc Generator
The^99 Mo–99mTc generator is constructed with alumina (Al 2 O 3 ) loaded in a
plastic or glass column. The^99 Mo activity is adsorbed on alumina in the
chemical form MoO 42 −(molybdate) and in various amounts. The amount of
alumina used is about 5–10 g depending on the^99 Mo activity. Currently, all
generators use fission-produced^99 Mo. The growth and decay of 99mTc along
with the decay of^99 Mo is shown in Figure 3.5 in Chapter 3. The 99mTc activ-
ity is eluted with 0.9% NaCl solution (isotonic saline) and obtained in the
chemical form of Na99mTcO4.
Considering that only 87% of^99 Mo decays to 99mTc, the 99mTc activity ATc
can be calculated from Eq. (3.15) as follows:
ATc=0.957(AMo) 0 (e−0.0105t−e−0.1155t) (5.4)where (AMo) 0 is the^99 Mo activity at t=0,lMo=0.0105 hr−^1 , and lTc=
0.1155 hr−^1. The timet has the unit of hour. At transient equilibrium [from
Eq. (3.16)],
ATc=0.957(AMo) 0 e−0.0105t
=0.957(AMo)t (5.5)Upon elution with saline, approximately 75% to 85% of the total activity
is eluted from the column. After about 4 half-lives, the 99mTc activity reaches
maximum.
Problem 5.1
A 2.5-Ci^99 Mo–99mTc generator calibrated for Thursday noon was received
on Wednesday afternoon. What would be the 99mTc activity, eluted at
8:00 a.m. on Friday, assuming a transient equilibrium between^99 Mo and
99mTc at the time of elution, and 85% elution yield?
Answer
(^99) Mo activity on Thursday noon =2500 mCi
The time from Thursday noon to 8:00 a.m. on Friday =20 hrs
Therefore,^99 Mo activity at 8:00 a.m. Friday = 2500 ×exp(−0.0105 ×20)
=2026 mCi
Using Eq. (5.5) at transient equilibrium,
99mTc activity =0.956 × 2026 ×0.85
=1646 mCi
The^99 Mo activity is likely to be eluted in trace quantities along with 99mTc
activity. This is called the^99 Mo or Moly breakthrough. The presence of^99 Mo
gives unnecessary radiation dose to the patient. According to the Nuclear
Radionuclide Generators 53