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Radioactive isotopes in use question help watch

The radioactive isotope of 131-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 to a patient in a glass of water. The isotope is required to have an activity of 800kBq at the time it is given to the patient.

Calculate:

A) the activity of the sample 24 hours after it

B) the activity of the sample when it was prepared 24 hours before it was given to the patient

C) the mass of the 131-I in the sample when it was prepared

I got the right answer for part A and B but I'm stuck on C. I know that Mass=moles x molar mass and that 1 mole = Avogadro constant but when I tried to do (131 x 6.02x10^23) I'm way off
2. (Original post by 221loki)

The radioactive isotope of 131-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 to a patient in a glass of water. The isotope is required to have an activity of 800kBq at the time it is given to the patient.

Calculate:

A) the activity of the sample 24 hours after it

B) the activity of the sample when it was prepared 24 hours before it was given to the patient

C) the mass of the 131-I in the sample when it was prepared

I got the right answer for part A and B but I'm stuck on C. I know that Mass=moles x molar mass and that 1 mole = Avogadro constant but when I tried to do (131 x 6.02x10^23) I'm way off
Activity A = λN

where N is the number you want in your equation.
You find λ from the half life and A is the activity when prepared if N is the number of active nuclei when prepared.
3. (Original post by Stonebridge)
Activity A = λN

where N is the number you want in your equation.
You find λ from the half life and A is the activity when prepared if N is the number of active nuclei when prepared.
so I found N then multiplied it by the Avogardo constant to get 6.92x10^11 then I did 131/)6.92x10^11) which is 1.8927x10^-10 g is that right? Or did I make up a load of rubbish?
4. (Original post by 221loki)
so I found N then multiplied it by the Avogardo constant to get 6.92x10^11 then I did 131/)6.92x10^11) which is 1.8927x10^-10 g is that right? Or did I make up a load of rubbish?
1. you need to take the value for the activity from part b) because it's about the time of preparation
2. N = A/λ give the number of atoms of iodine. (Have you got the correct λ?)

3. Then this number divided by the Avogadro constant gives the number of moles of iodine.

4. This then gives the mass.

That's the method.
5. Yes that's the correct answer, however I suspect you meant to say that you divided it by Avogadro's constant? Number of moles = N/NA
6. (Original post by Stonebridge)
1. you need to take the value for the activity from part b) because it's about the time of preparation
2. N = A/λ give the number of atoms of iodine. (Have you got the correct λ?)

3. Then this number divided by the Avogadro constant gives the number of moles of iodine.

4. This then gives the mass.

That's the method.
The activity from part B is 872193Bq so 870 kBq

λ is ln2/(8x24x60x60) = 1.00x10^-6

So N is 8.9674x10^11

8.9674x10^11 / 6.02x10^23 = 1.4448x10^-12

So mass is 131 x 1.4448x10^-12 = 1.89x10^-10 g = 1.9x10^-13kg

(Original post by james153)
Yes that's the correct answer, however I suspect you meant to say that you divided it by Avogadro's constant? Number of moles = N/NA
Yeah I think I've got it now thanks
7. (Original post by 221loki)
The activity from part B is 872193Bq so 870 kBq

λ is ln2/(8x24x60x60) = 1.00x10^-6

So N is 8.9674x10^11

8.9674x10^11 / 6.02x10^23 = 1.4448x10^-12

So mass is 131 x 1.4448x10^-12 = 1.89x10^-10 g = 1.9x10^-13kg

How do you get N is 8.9674x10^11

from N= A/λ

when A = 870kBq (so you say)
and λ = 1.00 x 106

Surely that gives N = 8.7 x 1011

The rest is correct.
8. (Original post by Stonebridge)
How do you get N is 8.9674x10^11

from N= A/λ

when A = 870kBq (so you say)
and λ = 1.00 x 106

Surely that gives N = 8.7 x 1011

The rest is correct.
I rounded A to 870kBq but the original figure was 872183Bq which is what I used to calculate N
I guess I wrote it wrong because now I calculated it again and got 8.722x10^11

Thanks again

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