A radioactive isotope of phosphorus, 32P, is used in internal radiotherapy to tr
ID: 1408281 • 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 1.36 MBq.
1. How many electrons are emitted over a period of 7.5 days?
2. What is the total amount of energy absorbed by the patients body during this period?
Eabsorbed =
3. If the radiation is absorbed by 121 g of tissue, what is the patient's equivient absorbed dosage?
Dose =
Explanation / Answer
1.
q = no. of P32 atoms, q0 = initial, qt = value at given time t (7.5 days, right?)
dq/dt(0) = 1300000
dq/dt(t) = dq/dt(0) * 2^(-t/Thalf) = 903315.293309448
e-based time constant TC = Thalf / ln(2) = 1.77998559988848D+06 s
q0 = -dq/dt(0)*TC = 2.31398127985502D+12 atoms
qt = q0 * 2^(-t/Thalf) = 1.60788821424985D+12 atoms
Change qt-q0 (electrons emitted) = 7.06093065605167D+11
2.
Energy absorbed = 7.06093065605167D+11 * 705 kEv
3.
Equivalent dosage = Energy absorbed*eVtoJ*RBE/mass (sieverts), or 100 * that number (rems)
(eVtoJ = 1/6.24150974E18)