Radioactivity Question

The radioactive isotope of 131/53 iodine is used for medical diagnosis of the kidneys. The isotope has a half life of 8 days. A sample of the isotope is to be given in a glass of water. The passage of the isotope through each kidney is then monitored using two detectors outside the body. The isotope is required to have an activity of 800 Bq at the time it is given to the patient.
A) Calculate:
I) The activity of the sample 24 hrs after it was given to the Patient.
II) The activity of the sample when it was prepared 24 hrs before it was given to the Patient.
III) The Mass of 131/53 Iodine in the sample when it was prepared
B) The readings from the detector near one of the patient kidneys rises then falls. The reading from the other detector which is near the other kidney rises and does not fall. Discuss the conclusion that can be drawn from these observations.
How do you do these questions. I have no clue.

(a)(i) - The activity immediately after it is given to the patient is 800 Bq. 24 hours (1 day) is 1/8 of a half life, so the fraction of the radioisotope left is (0.5)^(0.125) = 0.917..., so this should be the fraction of the activity after 1 day, e.g new activity = 800 Bq x 0.917... = 734 Bq (3 sf).
(a)(ii) - You could use a similar trick to part (a)(i). We have established that a day after being prepared, 91.7% of the sample remains and so if we call the old activity A, it stands to reason that the new activity (800 Bq) is given by 0.917A. Equating 0.917A and 800 Bq, then dividing both sides through by 0.917 gives A = 872 Bq (3 sf).
(a)(iii) - You need to use the equation A = λN (where λ is the decay constant and N is the number of particles of the radioisotope), where λ = ln(2)/T (where ln(2) has an approximate value of 0.693 and T is the half life of the isotope in seconds). Since we are given A (800 Bq = 800 s^-1) and T (albeit in days rather than seconds), we have to solve for N first (i.e A/λ = N). T = 8 days x 24 hours per day x 60 minutes per hour x 60 seconds per minute = 691200 seconds, so λ = 0.00000100281 s^-1. Therefore N = (800 s^-1)/(0.00000100281 s^-1) = 797752650 (nearest whole number). But this is the number of particles and not the mass of the sample. If you are studying A level chemistry (or equivalent), you will have come across mole calculations and will know how to proceed from here. In case you have not, it is worth learning that one mole of a substance corresponds to (roughly) 6.022 x 10^23 particles of said substance. That is to say that the sample contains 797752650/(6.022 x 10^23) = 1.32... x 10^-15 mol of iodine-131. The mass of the sample is given by the moles of particles times their molar mass (in this case 131 g/mol, as the mass of an isotope is its molar mass), so the mass of the sample = 1.32... x 10^-15 mol x 131 g/mol = 1.74 x 10^-13 g (3 sf). That is a surprisingly small amount.
(b) - At this point, I'm just guessing. The kidneys are involved in excretion, so if the reading rises then falls, the radioisotope passes though the kidney and is filtered into urine, which implies it is functioning normally. If the reading rises, but does not fall, then the radioisotope is passing into the kidney but not being filtered out and so the function of the kidney is abnormal or the filtration mechanism is obstructed. You will probably have to be more specific with your answer and state something like the patient has a kidney stone or a kidney tumour.

Thank you so much! Your a legend!
Only question I have is in part A, I which is where has the 0.5 come from again?

It’s because we are working with half lives. With each full half life, half the atoms decay - perhaps it would have been more obvious if I had written 1/2 instead of 0.5. As such, the fraction of atoms left = (1/2)^(number of half lives passed).