A radioactive isotope of phosphorus, ^32 p is used in internal radiotherapy to t

Question

A radioactive isotope of phosphorus, ^32 p is used in internal radiotherapy to treat patients with cancers and other diseases of the bone marrow. ^32 P has a half-life of 14.28 days and a atomic mass of ^32 P is 31.973 u. Via beta decay, it emits an electron with an energy of 705 KeV and a RBE of 14.5 As part of a treatment program, a patient ingests a radioactive pharmaceutical containing ^32 P which has an initial activity of 1.45 MBq. 1) How many electrons are emitted over a period of 7.5 days? N = electrons What is the total amount of energy absorbed by the patient's body during this period? E_absorbed = eV If the radiation Is absorbed by 123 g of tissue, what is the patients equivient absorbed dosage? Dose = Rem

Explanation / Answer

a)We need to know the decay constant first.

lambda = ln 2/T(1/2)

lambda = ln 2/(14.28 days x 86400 s/ 1 day) = 5.61 x 10^-7 per sec

The number of nuclei intially is:

N0 = R0/lambda

N0 = 1.45 x 10^6 Bq/5.61 x 10^-7 = 2.58 x 10^12 Nuclei

The number of nuclei after some time t is given by:

N = N0 e^-lambda x t

t = 7.5 days = 7.5 x 86400 s = 648000 s

N = 2.58 x 10^12 x e^-(5.61 x 10^-7 x 648000) = 1.79 x 10^12 Nuclei

So the number of emitted electrons are:

n = N0 - N = (2.58 - 1.79) x 10^-12 = 7.9 x 10^11 electrons

Hence, n = 7.9 x 10^11 electrons

b)absorbed energy is:

E = n E

E = 7.9 x 10^11 x 705 x 10^3 eV = 5.57 x 10^17 eV

E = 5.57 x 10^17 eV

c)Dosage will be:

Dose = 5.57 x 10^17 eV x 1.6 x 10^-19/0.123 kg = 0.7246 x 100 Rad = 72.46 rad

In rem,

Dose = 72.46 x rad(15) = 1086.9 rem

Hence, Dose = 1086.9 Rem

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