Answer to Question #120076 in Physics for Raghuveer Reddy

Question

3. A 14 g ovarian tumor is treated using a sodium phosphate solution in which the phosphorus atoms are the
radioactive 32P isotope with a half-life of 14.3 days and which decays via beta emission with an energy of
1.71MeV. Half of the sodium phosphate solution is absorbed by the tumor and deposits 10 J of energy into it.
The other half of the solution is dispersed throughout the patients’ tissues, also depositing 9 J of energy into the
50.0 kg of body tissues. (a) What is the dose (in Gy and rem) that the tumor receives? (b) What is the dose (in Gy
and rem) that the rest of the patient receives?

Answer 1

By definition, the dose absorbed by the body with mass $m$ after receiving energy $E$ is the following:

$$D = E/m$$

Thus, the tumor obtaind a dose:

$$D_{tumor} = 10J/0.014kg = 714.3 Gy$$

The body obtained the dose:

$$D_{body} = 9J/50kg = 0.18 Gy$$

To convert these number to rem one should multiply them by 100.

Answer. The tumor receives 714.3 Gy or 71430 rem. The rest of the patient receives 0.18 Gy or 18 rem.