Nuclear medicine, total energy, dose in rem

[NickMcCartney]

I can't seem to get this started in the right direction. Any help is appreciated.

As part of a treatment program, a patient ingests a radioactive pharmaceutical containing P32,15, which emits beta rays with an of 1.50. The half-life of is 14.28 , and the initial activity of the medication is 1.34 .

Part A
How many electrons are emitted over the period of 7.00 days?

Part B
If the rays have an energy of 705 keV , what is the total amount of energy absorbed by the patient's body in 7.00 ?

Part C
Find the absorbed dosage in rem, assuming the radiation is absorbed by 110 grams of tissue.

First I converted all times to seconds.

T 1/2 = 1.234e6 s
t = 7 days = 604800 s

Converted activity (RO) into decays per second = 1.34e6

Found N from A-Z = 17

Found decay constant (lambda) = ln(2)/T 1/2 = 5.617e-7

Equations I have tried with no success are N=Noe^-(lambda)(t)

 

[NateD]

and A=Aoe^-(lambda)(t)Part A:To find the number of electrons emitted over the period of 7.00 days, we can use the equation N=Noe^-(lambda)(t), where N is the number of electrons at time t, No is the initial number of electrons, and lambda is the decay constant (in this case, 5.617e-7). Plugging in the values we have, we get N = 1.34e6 * e^-(5.617e-7 * 604800) = 1.091e6 electrons.Part B: To find the total amount of energy absorbed by the patient's body in 7.00 days, we can use the equation E = N * Eo, where E is the total energy absorbed, N is the number of electrons emitted (1.091e6), and Eo is the energy for each electron (705 keV). Plugging in the values we have, we get E = 1.091e6 * 705 keV = 763.8 MeV.Part C:To find the absorbed dosage in rem, we can use the equation D = E * 10^-3 / (N * A), where D is the absorbed dosage, E is the total energy absorbed (763.8 MeV), N is the number of electrons emitted (1.091e6), and A is the atomic mass of the tissue (110 g/mol). Plugging in the values we have, we get D = 763.8 * 10^-3 / (1.091e6 * 110) = 0.0693 rem.