A radioactive isotope of phosphorus, 32P, is used in internal radiotherapy to tr
ID: 1300095 • Letter: A
Question
A radioactive isotope of phosphorus, 32P, is used in internal radiotherapy to treat patients with cancers and other diseases of the bone marrow. 32P has a half-life of 14.28 days and a atomic mass of 32P is 31.973 u. Via ? decay, it emits an electron with an energy of 705 KeV and a RBE of 15 . As part of a treatment program, a patient ingests a radioactive pharmaceutical containing 32P which has an initial activity of 1.39 MBq.
1) How many electrons are emitted over a period of 6 days?
N = __________electrons
2) What is the total amount of energy absorbed by the patients body during this period?
Eabsorbed = ________eV
3) If the radiation is absorbed by 123 g of tissue, what is the patient's equivient absorbed dosage?
Dose = ___________Rem
Explanation / Answer
Decay constant = 0.693 / 14.28 = 5.618e-7
( NOTE - convert 14.28 days to seconds by 14.28*24*60*60 = 1233792 seconds)
(a) No = 1.39 / 5.618e-7 = 2.474e12
Therefore, electrons emitted over a period of 6 days is given as
N = No e-decayconstant * t
N = 2.474e12 e-5.618e-7 * 6*24*3600
N = 2.474e12 e-0.29123712
N = 1.8489e12
Thereofore, Number of electrons emitted = 2.474e12 - 1.8489e12
Number of electrons emitted = 6.25e11
(b) total amount of energy is given as
E = 6.25e11 * 705e3
E = 4.4e17 eV
(c) Absorbed dosage
(4.4e17 * 1.6e-19 / 0.123) * 100 = 57.31
dosage = 57.235*15 = 858.5 Rem