average_human
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I'm currently studying half-life in physics for GCSE and I came across this question on Isaac Physics. I've tried it multiple times, and I keep getting it wrong:
'The half-life of 13 7 N is 10 minutes. The initial activity of a sample of
13 7 N is 100Bq. Determine the half-life 5.0 minutes later'.

Could anyone help me out on this? thanks
Also, the 13 is supposed to be at the top of the N, and the 7 is at the bottom, if that makes sense x
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MagnusStormrider
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What are the answers you've got? And is that the exact wording of the question? Half-life for a given isotope is the same, no matter how much time has passed. Is the question asking for the activity of the sample after five minutes?
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average_human
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(Original post by MagnusStormrider)
What are the answers you've got? And is that the exact wording of the question? Half-life for a given isotope is the same, no matter how much time has passed. Is the question asking for the activity of the sample after five minutes?
yes, so it's like half of the half life i think
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MagnusStormrider
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Some definitions:
Half-life- the time it takes for the activity of a radioactive sample to fall by half.
Activity- the amount of radiation emitted by a radioactive sample in a given time-frame. 1 Bequerel is equivalent to one unstable nucleus decaying per second.
The question asks for the activity of the sample after five minutes. This does not mean anything like the 'half-life of the half-life'. Activity is distinct from half-life, but knowing the half-life and initial activity allows you to calculate the activity at a given time.
So, to calculate the activity after any given length of time, you first have to determine how many half-lives have passed. Do you know how to work that out?
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average_human
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(Original post by MagnusStormrider)
Some definitions:
Half-life- the time it takes for the activity of a radioactive sample to fall by half.
Activity- the amount of radiation emitted by a radioactive sample in a given time-frame. 1 Bequerel is equivalent to one unstable nucleus decaying per second.
The question asks for the activity of the sample after five minutes. This does not mean anything like the 'half-life of the half-life'. Activity is distinct from half-life, but knowing the half-life and initial activity allows you to calculate the activity at a given time.
So, to calculate the activity after any given length of time, you first have to determine how many half-lives have passed. Do you know how to work that out?
no, I don't. thanks for clarifying the definitions as well
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Arthur_Morgan
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The title got me excited for a moment there
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MagnusStormrider
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(Original post by average_human)
no, I don't. thanks for clarifying the definitions as well
Have you done any questions on binary fission and mean division time in biology? If you have, it's a bit like that. If not, don't worry.
The question says that the half-life is 10 minutes, and you have to work out the activity after five minutes. This is an odd question, because usually the given half-life is shorter than the time which has passed.
What you have to do is to work out how many times the half-life time has passed. In other words, how many times does the half-life occur within the given period.

In this case, 5/10=0.5, so 0.5 half-lives have occurred.
If one half-life had passed, then the initial activity would have fallen by half, but since only half a half-life has passed, the activity has dropped by a quarter (meaning that the activity is now 3/4 what it was originally, not 1/4). You can work out the answer from this and the initial activity.

Another example:
If, say, the question asked for the activity after 20 minutes, and the half-life was still 10 minutes, the method is the same. 20/10=2.
In this case, the activity would have halved twice. With an initial activity of 100Bq:
After first half-life: 100/2= 50Bq
After second half-life: 50/2= 25Bq
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average_human
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(Original post by MagnusStormrider)
Have you done any questions on binary fission and mean division time in biology? If you have, it's a bit like that. If not, don't worry.
The question says that the half-life is 10 minutes, and you have to work out the activity after five minutes. This is an odd question, because usually the given half-life is shorter than the time which has passed.
What you have to do is to work out how many times the half-life time has passed. In other words, how many times does the half-life occur within the given period.

In this case, 5/10=0.5, so 0.5 half-lives have occurred.
If one half-life had passed, then the initial activity would have fallen by half, but since only half a half-life has passed, the activity has dropped by a quarter (meaning that the activity is now 3/4 what it was originally, not 1/4). You can work out the answer from this and the initial activity.

Another example:
If, say, the question asked for the activity after 20 minutes, and the half-life was still 10 minutes, the method is the same. 20/10=2.
In this case, the activity would have halved twice. With an initial activity of 100Bq:
After first half-life: 100/2= 50Bq
After second half-life: 50/2= 25Bq
i worked out 3/4 and got 75Bq but it said i was wrong
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MagnusStormrider
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Do you have the answers, or is it an online thing which just tells you if you're right or wrong?
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average_human
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(Original post by MagnusStormrider)
Do you have the answers, or is it an online thing which just tells you if you're right or wrong?
just an online thing. this is what it said when i put in 75Bq:

Because the decay is exponential rather than linear, the activity after half of a half-life is not 3/4 of the initial activity.
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MagnusStormrider
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That makes sense, though with that, I don't see how to find the answer without drawing a graph, sorry.
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average_human
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(Original post by MagnusStormrider)
That makes sense, though with that, I don't see how to find the answer without drawing a graph, sorry.
i got the answer. it was 71. thanks for your help with the definitions
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MagnusStormrider
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How? Did you guess, or find a method?
And no problem
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average_human
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(Original post by MagnusStormrider)
How? Did you guess, or find a method?
And no problem
there was some formula where you had to do 2 to the power of n (0.5). Then divide 100Bq by the answer
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