TheoP31
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"A sample of ancient wood of mass 0.50 g is found to have an activity of 0.11 Bq. A sample of living wood of the same mass has an activity of 0.13 Bq. Calculate the age of the sample of wood.
The half life of radioactive carbon-14 is 5570 years."

N.B. the term "living wood" is used to imply that the wood is radioactive.


The answer to this question is: 1340 years, but I am not sure how to get to this answer.


I've tried several methods of working out but nothing seems to be anywhere close to this answer.

Any help is greatly appreciated.
Thank you.
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Joshwoods01
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have you tried using the equation for half life? I assume you have but worth asking haha
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TheoP31
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(Original post by Joshwoods01)
have you tried using the equation for half life? I assume you have but worth asking haha
Yes, I tried using the half life equation, to work out the decay constant. And then I used that decay constant in the activity equation to work out N.
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RayyanT
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(Original post by TheoP31)
"A sample of ancient wood of mass 0.50 g is found to have an activity of 0.11 Bq. A sample of living wood of the same mass has an activity of 0.13 Bq. Calculate the age of the sample of wood.
The half life of radioactive carbon-14 is 5570 years."

N.B. the term "living wood" is used to imply that the wood is radioactive.


The answer to this question is: 1340 years, but I am not sure how to get to this answer.


I've tried several methods of working out but nothing seems to be anywhere close to this answer.

Any help is greatly appreciated.
Thank you.
First use the equation A=Ao x e^- lambda x t (make sure there is a minus throughout the whole equation)
After plugging in numbers you get
0.11=0.13 x e^- lamdba x t (t is what we are trying to find. So re arrange the equation. Divide 0.11/0.13, and then take the logs of both sides so after re arranging we are left with):
ln(0.11/0/13)= - lamdba x t
Then to make t the formula we have to divide the left hand side by -lambda. So the final main equation is:

ln(0.11 divided by 0.13) all divided by -lamdba = t

So the last thing we need to do is find out lambda.

Lambda= ln2 / t^1/2 (t^1/2 is known as the half life which is given as 5570 YEARS. CONVERT TO SECONDS.)

After plugging into the equation we get:
ln2 / 5570 x 365 x 24 x 60 x 60= 3.946 e-12

So we now know lambda.

So back to our original final equation:

ln(0.11/0.13) divided by MINUS 3.946e-12

You should get an answer of 4.23344e10. This is the answer but in seconds. To get it into years all we have to do is divide by 60, divide by 60, divide by 24, divide by 365 and that should give you the answer.

You better like this cus i put time into this sh1t
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TheoP31
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(Original post by RayyanT)
First use the equation A=Ao x e^- lambda x t (make sure there is a minus throughout the whole equation)
After plugging in numbers you get
0.11=0.13 x e^- lamdba x t (t is what we are trying to find. So re arrange the equation. Divide 0.11/0.13, and then take the logs of both sides so after re arranging we are left with):
ln(0.11/0/13)= - lamdba x t
Then to make t the formula we have to divide the left hand side by -lambda. So the final main equation is:

ln(0.11 divided by 0.13) all divided by -lamdba = t

So the last thing we need to do is find out lambda.

Lambda= ln2 / t^1/2 (t^1/2 is known as the half life which is given as 5570 YEARS. CONVERT TO SECONDS.)

After plugging into the equation we get:
ln2 / 5570 x 365 x 24 x 60 x 60= 3.946 e-12

So we now know lambda.

So back to our original final equation:

ln(0.11/0.13) divided by MINUS 3.946e-12

You should get an answer of 4.23344e10. This is the answer but in seconds. To get it into years all we have to do is divide by 60, divide by 60, divide by 24, divide by 365 and that should give you the answer.

You better like this cus i put time into this sh1t
Thank you so much Rayyan! That makes sense now.
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RayyanT
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(Original post by TheoP31)
Thank you so much Rayyan! That makes sense now.
nws if you have any more Qs lmk- revised most topics for mocks next week
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TheoP31
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(Original post by RayyanT)
nws if you have any more Qs lmk- revised most topics for mocks next week
Just thought I'd mention - you do not need to convert the half-life into seconds. It'll save working out in the exam by keeping it in hours, so you don't have to convert back afterwards.
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RayyanT
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(Original post by TheoP31)
Just thought I'd mention - you do not need to convert the half-life into seconds. It'll save working out in the exam by keeping it in hours, so you don't have to convert back afterwards.
oh fair enough true true
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TheoP31
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RayyanT
I have another one which I'm stuck on if you want to give it a go, for practise.

"The radioactive isotope of iodine, 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 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 kBq at the time it is given to the patient."

"Calculate:
i) the activity of the sample 24 hours after it was given to the patient,
ii) the activity of the sample when it was prepared 24 hours before it was given to the patient,
iii) the mass of 131-Iodine in the sample when it was prepared."


I have managed to work out the answers for part 1 and part 2. And they're both correct.
i) 734 kBq
ii) 872 kBq

But I am not sure about part 3 of this question. I have tried several methods involving using Avogadro's constant and mass = moles x Mr formula, but none seem to get me the correct answer.

The answer to this question should be: 1.9x10^-13 kg.

Any help is much appreciated Sir.
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RayyanT
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(Original post by TheoP31)
RayyanT
I have another one which I'm stuck on if you want to give it a go, for practise.

"The radioactive isotope of iodine, 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 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 kBq at the time it is given to the patient."

"Calculate:
i) the activity of the sample 24 hours after it was given to the patient,
ii) the activity of the sample when it was prepared 24 hours before it was given to the patient,
iii) the mass of 131-Iodine in the sample when it was prepared."


I have managed to work out the answers for part 1 and part 2. And they're both correct.
i) 734 kBq
ii) 872 kBq

But I am not sure about part 3 of this question. I have tried several methods involving using Avogadro's constant and mass = moles x Mr formula, but none seem to get me the correct answer.

The answer to this question should be: 1.9x10^-13 kg.

Any help is much appreciated Sir.
I’ll give it a go tomorrow bro
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TheoP31
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(Original post by RayyanT)
I’ll give it a go tomorrow bro
Thank you mate! Nuclear physics is definitely your 'forte' haha.
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RayyanT
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(Original post by TheoP31)
RayyanT
I have another one which I'm stuck on if you want to give it a go, for practise.

"The radioactive isotope of iodine, 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 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 kBq at the time it is given to the patient."

"Calculate:
i) the activity of the sample 24 hours after it was given to the patient,
ii) the activity of the sample when it was prepared 24 hours before it was given to the patient,
iii) the mass of 131-Iodine in the sample when it was prepared."


I have managed to work out the answers for part 1 and part 2. And they're both correct.
i) 734 kBq
ii) 872 kBq

But I am not sure about part 3 of this question. I have tried several methods involving using Avogadro's constant and mass = moles x Mr formula, but none seem to get me the correct answer.

The answer to this question should be: 1.9x10^-13 kg.

Any help is much appreciated Sir.
So we know the activity 24 hours before because that's the answer to part B.

use the equation A=lambda x N, rearrange to make N the formula. A is the answer to B and lambda is ln2/ the half life.
when we sub in values a is 872 e3 and lambda is ln2/8 x 24 x 60 x 60 (1)

N= m x Avogadro's constant / Ar number.

So I plugged it in. N is A/Lambda, m we're trying to work out. Avogadro's is 6.02e23 all multiplied by the Ar number which is 131.
so the answer to (1) x 131. This answer is then divided by avogadro's constant, which gives us mass.

and thats the answer. lmk if u need anymore help, my number is 07827441315 if u want any more questions or u can post them here idrm doing them good practise.

The worst thing is my schools considering taking the whole or most of nuclear physics out lmao the only thing i revised...
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TheoP31
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(Original post by RayyanT)
So we know the activity 24 hours before because that's the answer to part B.

use the equation A=lambda x N, rearrange to make N the formula. A is the answer to B and lambda is ln2/ the half life.
when we sub in values a is 872 e3 and lambda is ln2/8 x 24 x 60 x 60 (1)

N= m x Avogadro's constant / Ar number.

So I plugged it in. N is A/Lambda, m we're trying to work out. Avogadro's is 6.02e23 all multiplied by the Ar number which is 131.
so the answer to (1) x 131. This answer is then divided by avogadro's constant, which gives us mass.

and thats the answer. lmk if u need anymore help, my number is 07827441315 if u want any more questions or u can post them here idrm doing them good practise.

The worst thing is my schools considering taking the whole or most of nuclear physics out lmao the only thing i revised...
Thank you so much! That helps a lot!

They're really taking it out of your exams? DAMN. Maybe because you covered it in lockdown or smth?
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RayyanT
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Nws and Yh covered it in lockdown haven't decided anything yet but it’s that or astrophysics I think. If you don’t mind me asking have u applied to uni and if so what course?
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