Question

The quantity of a radioactive material is often measured by its activity (measured in curies or millicuries) rather than by its mass. In a brain scan procedure, a 84.0−kg patient is injected with 19.0 mCi of 99mTc, which decays by emitting γ−ray photons with a half-life of 6.01 h. Given that the RBE of these photons is 0.98 and exactly two-thirds of the photons are absorbed by the body, calculate the rem dose received by the patient. Assume all the 99mTc nuclei decay while in the body. The energy of a γ−ray photon is 2.29 × 10−14 J

Answer 1

Activity of Tc-99 injected =Ro = 19.0 mCi = 19.0 x 10^-3 Ci

We know 1 Ci = 3.70x10¹⁰ decays/s

Activity of Tc-99 in decays/s =

=70.3 x 10⁷ decays/s

Ro=7.03 x 10⁸ decays/s

The formula for activity =Ro ,

where activity constant and

No = number of Tc-99 nuclei present

Activity constant =

Given, half-life time = 6.01 h =

=21,636 s

Plugin the values and finding activity in seconds =

Now use this values and solve for No

No = 2.195 x 10¹³ decays

each nuclei emits one photon so

number of decays = number of nuclei =  2.195 x 10¹³ decays

The energy of one photon = 2.29 × 10⁻¹⁴ J

The total energy of the absorbed phtons = two-thirds of the all photons

E=0.335 J

Converting energy to rad =

=0.399 rad

Given that RBE = 0.98

then rem dose received by the patient = 0.98(0.399) = 0.391 rem