Alen.m
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Please could someone give me a hint or help me see what I could possibly have done wrong with this question?
"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 was given to the patient (I got the correct answer to this which is 730kBq)
b) the activity of the sample when it was prepared 24 hours before it was given to the patient ( I also managed to get the correct answer to this which is 870kBq)
c) the mass of the 131-I in the sample when it was prepared (for this part i tried to work out the constant using A=CONSTANT X N after this I'm absolutely lost how to find out mass
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Plagioclase
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(Original post by Alen.m)
Please could someone give me a hint or help me see what I could possibly have done wrong with this question?
"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 was given to the patient (I got the correct answer to this which is 730kBq)
b) the activity of the sample when it was prepared 24 hours before it was given to the patient ( I also managed to get the correct answer to this which is 870kBq)
c) the mass of the 131-I in the sample when it was prepared (for this part i tried to work out the constant using A=CONSTANT X N after this I'm absolutely lost how to find out mass
You've got the right formula, so that's a start. Have you managed to work out N yet?
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Alen.m
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well actually that's my problem, i consider N as a Avagado's number but then i realised it gives me a wrong answer so there would be another way to calculate N. Is it the formula N=N(0) x e^decay constant x t (where t=24 hrs and N(0) is equal to Avagado number?)
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Plagioclase
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(Original post by Alen.m)
well actually that's my problem, i consider N as a Avagado's number but then i realised it gives me a wrong answer so there would be another way to calculate N. Is it the formula N=N(0) x e^decay constant x t (where t=24 hrs and N(0) is equal to Avagado number?)
N is not Avogadro's number!! In the equation A=\lambda N, N refers to the number of radioactive nuclei. This is the correct formula to use and Avogradro's number will be needed later in the question but first of all you need to work out N (using this formula).

(please quote me in so I know when you've replied)
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Alen.m
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i've calculated N using formula above and then switch into the formula N=(m/M)x N(a), where N(a) is Avogadro's number , M is the mass number which is 131 and N has been calculated as perviously .the answer i get for m is 1.1x10^-14 Kg but the text book says the answer is 1.9 x 10*-13 Kg .
hope I'm in the right track .
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Plagioclase
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(Original post by Alen.m)
i've calculated N using formula above and then switch into the formula N=(m/M)x N(a), where N(a) is Avogadro's number , M is the mass number which is 131 and N has been calculated as perviously .the answer i get for m is 1.1x10^-14 Kg but the text book says the answer is 1.9 x 10*-13 Kg .
hope I'm in the right track .
Please quote me when you reply (using the reply button) otherwise I won't know you have replied and it will take me longer to respond.

Could you please post your working, because I have also got 1.9 x 10-13kg. Specifically, what did you get as the decay constant? Did you remember that the half life is 8 days?
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Absent Agent
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Alen.m
You man never quote
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Alen.m
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(Original post by Plagioclase)
Please quote me when you reply (using the reply button) otherwise I won't know you have replied and it will take me longer to respond.

Could you please post your working, because I have also got 1.9 x 10-13kg. Specifically, what did you get as the decay constant? Did you remember that the half life is 8 days?
sorry i guess i've been revising the whole day so my brain kinda stop functioning well which is why i keep forgetting things
i've attached my full working here for you hope you can help me spot my problem)
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Alen.m
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(Original post by Mehrdad jafari)
Alen.m
You man never quote
working on it mate
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Plagioclase
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(Original post by Alen.m)
sorry i guess i've been revising the whole day so my brain kinda stop functioning well which is why i keep forgetting things
i've attached my full working here for you hope you can help me spot my problem)
Thanks for quoting me. And you may want to check your calculation for the decay constant because it's not 1.6x10-6. Remember that \lambda=\frac{ln2}{\tau} where \tau is in seconds.
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Alen.m
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(Original post by Plagioclase)
Thanks for quoting me. And you may want to check your calculation for the decay constant because it's not 1.6x10-6. Remember that \lambda=\frac{ln2}{\tau} where \tau is in seconds.
no problem mate
i've just checked my calculation . i've also had T in seconds which was 691200 sec(8days) so 0.693/691200 would give me the decay constant which i got it for 1x10^-6 . i also used it to calculate the activity of the sample which i got it for 869565 so i assumed it's been rounded up to 870000 and that's why the book says 870000 as the answer for part b. so i really don't know what part I'm getting wrong in part c would you be able to post your working so i can spot my problem through yours?
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Plagioclase
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(Original post by Alen.m)
no problem mate
i've just checked my calculation . i've also had T in seconds which was 691200 sec(8days) so 0.693/691200 would give me the decay constant which i got it for 1x10^-6 . i also used it to calculate the activity of the sample which i got it for 869565 so i assumed it's been rounded up to 870000 and that's why the book says 870000 as the answer for part b. so i really don't know what part I'm getting wrong in part c would you be able to post your working so i can spot my problem through yours?
\lambda=\frac{ln2}{60 \times60 \times24 \times8}=1.00 \times10^{-6}
N=\frac{A}{\lambda}=\frac{8.72 \times10^5}{1.00 \times10^{-6}}=8.72 \times10^{11}
n=\frac{N}{N_A}=\frac{8.72 \times10^{11}}{6.02 \times10^{23}}=1.44 \times10^{-12}
M=n \times A_r=1.44 \times10^{-12} \times131=1.90 \times10^{-10}g=1.90 \times10^{-13}kg

Where did you go wrong?

(I've corrected the terrible formatting, sorry if you saw this before I corrected it!)
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Alen.m
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(Original post by Plagioclase)
\lambda=\frac{ln2}{60 \times60 \times24 \times8}=1.00 \times10^{-6}
N=\frac{A}{\lambda}=\frac{8.72 \times10^5}{1.00 \times10^{-6}}=8.72 \times10^{11}
n=\frac{N}{N_A}=\frac{8.72 \times10^{11}}{6.02 \times10^{23}}=1.44 \times10^{-12}
M=n \timesA_r=1.44 \times10^{-12} \times131=1.90 \times10^{-10}g=1.90 \times10^{-13}kg

Where did you go wrong?

(I've corrected the terrible formatting, sorry if you saw this before I corrected it!)
I guess the formula you use to sort out the problem is a bit different than mine especially the last part so im gonna have to go through it all over again using your formula but thanks to for your effort mate .
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Plagioclase
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(Original post by Alen.m)
I guess the formula you use to sort out the problem is a bit different than mine especially the last part so im gonna have to go through it all over again using your formula but thanks to for your effort mate .
Pretty sure that your formula basically does exactly what I've done, I just did each step individually. Hope it works out, if you need more help then just quote me in
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Alen.m
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(Original post by Plagioclase)
Pretty sure that your formula basically does exactly what I've done, I just did each step individually. Hope it works out, if you need more help then just quote me in
just a quick question on the last two parts. what does n(r) stand for and if you multiply that by M will give you m?im also not familiar with n=N/N(r)
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Plagioclase
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(Original post by Alen.m)
just a quick question on the last two parts. what does n(r) stand for and if you multiply that by M will give you m?im also not familiar with n=N/N(r)
Oh sorry, that's my terrible LaTeX skills (I've corrected it now, not sure how that happened). It's meant to say M (mass) = n (number of moles) x Ar (relative atomic mass).

n (number of moles) is N (number of particles) divided by avogadro's number (NA). You had this all in your formula, I just did it in two steps (step one to work out the number of moles, step two to work out the mass).
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Alen.m
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(Original post by Plagioclase)
Oh sorry, that's my terrible LaTeX skills (I've corrected it now, not sure how that happened). It's meant to say M (mass) = n (number of moles) x Ar (relative atomic mass).

n (number of moles) is N (number of particles) divided by avogadro's number (NA). You had this all in your formula, I just did it in two steps (step one to work out the number of moles, step two to work out the mass).
well i guess both our ways are correct. i've checked all my calculations again today and there was nothing wrong with it but my answer is still off by the the factor of 10^-1. thanks for your help anyway.
here's another question i'd like to ask if you can sort it out :
the reading from the detector near 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 conclusions that can be drawn from these observations? Does it have something to do with the foil that causes the reading to drop and rises?simply have no idea how to explain it
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Plagioclase
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(Original post by Alen.m)
the reading from the detector near 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 conclusions that can be drawn from these observations? Does it have something to do with the foil that causes the reading to drop and rises?simply have no idea how to explain it
Is that the whole question? Exactly as it's written in the book or wherever you're getting it from? You're talking about a foil but the question doesn't mention foil...
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Alen.m
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(Original post by Plagioclase)
Is that the whole question? Exactly as it's written in the book or wherever you're getting it from? You're talking about a foil but the question doesn't mention foil...
Actually yeah it is the whole question and the reason i've mentioned foil is because on the text book says the reading from the dector depends on the thickness of the foil. Either im confusing it with something else or im in the right path i dont know but your help would be much appreciated .
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Plagioclase
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(Original post by Alen.m)
Actually yeah it is the whole question and the reason i've mentioned foil is because on the text book says the reading from the dector depends on the thickness of the foil. Either im confusing it with something else or im in the right path i dont know but your help would be much appreciated .
I honestly don't know the answer to this - if I were forced to make a guess then I'd say the first kidney isn't working properly because the reading falls, suggesting there's not a lot of I-133 in it which suggests that fluids aren't flowing through it properly (whereas the reading of the other kidney doesn't fall, suggesting there is I-133 inside it which suggests that stuff is going through it normally) but I don't understand why the reading of the first kidney goes up in the first place.
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