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Easy half life question

  • Thread starter joejo
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150
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hey can someone help me out with this question...I have an test 2mr on this time and dont have any more time to waste. So can someone please show me how to do it...so I can study of it....thanks in advance


Scientist are able to determine the age of very old, naturally occuring items through the process of carbon dating. Carbon dating compares the amount of carbon-14 (C-14) that is present in an item with the amount that would be present if the item were created in the present. The half-life of C-14 is approxiamtely 6000 years.


a) Suppose that a piece of wood is cut from a tree. The piece of wood contains 32g of C-14. Write an equation to determine the amount of C-14 remaining after n half-lives.

b) Determine the amount of C-14 that would remain after 12000 years.

c) Suppose that scientists in the future carefully examine the piece of woor and determine that it contains 2g of C-14. How far in the future has the wood survived? Explain your solution.
 
788
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The basic form of exponential growth/decay, is

[tex]y = Ce^{kt}[/tex] .

k is positive or negative depending on if there is growth or decay. Start from there.
 

mrjeffy321

Science Advisor
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Each half life, there is half of the amount of what there was before of the substance experiencing the half life (in this case carbon 14).
Originally there was 32 grams, after 1 half life (6000 years), 16 grams, then another 600 year later, 8 grams left, ....

This process works like exponential decay,
original amount * .5^n, where n is the number of half-lives it has had.

to find out the number of half-lives it has experienced, simply divide the total time by the time for 1 half life,
12000 years over 6000 year = number of half lives

for the last question it it a bit trickier, it involve logorithms, but is not too hard if you remember the formula of how to detemine how much of a substance is left,
32 grams * .5^n = 2 grams
divide both sides by 32 to get,
.5^n = 1/16
then solve for n,
log(.5^n) = log(1/16), which =,
n*log(.5) = log(1/16)
divide both sides by log(.5) to get n,
then check the answer to make sure it it right.
 
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we haven't done logs....this is suppose to be a remedial math...I'm looking for someone to answer it properly so I use this question to study from....Can someone please help me out...
 

Doc Al

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joejo said:
we haven't done logs....this is suppose to be a remedial math...I'm looking for someone to answer it properly so I use this question to study from....Can someone please help me out...
You don't need logs to do part c. Just count how many times you need to divide 32g in half to end up with 2g--that will tell you the number of half-lives that must have passed.
 
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im lost...can someone show me how to do it...i need to study of this now....
 
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a) general form [itex]a \cdot b^n[/itex], so [itex]32(\frac{1}{2})^{\frac{n}{6000}}[/itex] - "n" is years in this problem

b) substitute for n

c) trial and error without logs :smile:
 
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still lost....i dont get it....im trying to do this correspondance and its very difficult
 
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Lets start with what a half-life is- how long it takes until a substance has lost half its mass.
Therefore, in one half life (in this case 6000 years), half of the mass of the substance will be ‘lost.’ Thus, after 6000 years, only 16g (32/2) will remain.
In another 6000 years (after 2 half-lives), half of this amount will be ‘lost’ leaving 8g (16/2) of the substance.
After the 3rd half-life, half this will be ‘lost’ leaving 4g. And so on.

You should be able to calculate the equation from here for part a.


For part b, 12000 years is equal to 2 half lives. You can use your equation from part a to calculate how much is left after 2 half lives.


For c, what value of n in your initial equation gives the value 2g? Each half life is equal to 6000 years so you can convert ‘half-lives’ to years.
 
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so what would the equation be?
 
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I'm not sure in this forum that we can just give you the equation but you should be able to work it out form the imformation given. This table should make it clearer.

Number of half-lives
Mass of Carbon-14 (g)​
0
32 (32/1)​
1
16 (32/2)​
2
8 (32/4)​
3
4 (32/8)​
4
2 (32/16)​
5
1 (32/32)​
etc
etc​


Note that to find the mass of carbon, the 32 is divided by roots of 2 (as shown). This should be reflected in your equation.

Show your working towards finding the equation and we can see where you are stuck/have gone wrong.
 
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is that it?

[itex]32(\frac{1}{2})^{\frac{n}{6000}}[/itex]
 

HallsofIvy

Science Advisor
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Is that what? Your problem had 3 parts to it. Which one does that answer? And we won't be able to tell you if it answers any one of those until you tells us what n means!
 
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hallsofIvy you never seem to help people out....


anyways...n is the number of half-lives it has had...is that right Doc G or anyone else that is willing to help...
 
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joejo said:
hallsofIvy you never seem to help people out....
You're wrong. HallsofIvy always helps people out, and he's trying to get you to learn something instead of spoon-feeding you the answer.

anyways...n is the number of half-lives it has had...is that right Doc G or anyone else that is willing to help...
"n" in multiples of 6000 would equal the number of half-lives gone by in the equation you posted, so what does that tell you about what n represents?
 
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the number of half lives....my test is 2mr...can someone plz answer it...its almost 2am and i dont get half life questions
 

HallsofIvy

Science Advisor
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All I can do is repeat what people have already told you:
a) Suppose that a piece of wood is cut from a tree. The piece of wood contains 32g of C-14. Write an equation to determine the amount of C-14 remaining after n half-lives.
"half" life means exactly that- the time required for the amount to halve. To find the amount remaining after n half-lives, you multiply by 1/2 n times.
The equation is just 32(1/2)n.

b) Determine the amount of C-14 that would remain after 12000 years.
You are told that the half-life of C-14 is about 6000 years. 12000 is two half-lives.
Multiply by 1/2 twice (0r 32(1/2)2)

c) Suppose that scientists in the future carefully examine the piece of woor and determine that it contains 2g of C-14. How far in the future has the wood survived? Explain your solution.
The piece of wood starts with 32 g and eventually has only 2 g. It has 2/32= 1/16= (1/2)4 as much as originally. That means 32 is multiplied by 1/2 4 times. How many half-lives is that? How many years?
 
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thanks hall of ivy....as for that below...

there are 4 half lives...making it 8 years correct?



The piece of wood starts with 32 g and eventually has only 2 g. It has 2/32= 1/16= (1/2)4 as much as originally. That means 32 is multiplied by 1/2 4 times. How many half-lives is that? How many years?[/QUOTE]
 

Doc Al

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joejo said:
there are 4 half lives...making it 8 years correct?
No. How long is one half-life?
 
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375 years......my test is in 5 hours...can someone please just check my answer and if its wrong state the correct answer...i've been up all night...thanks
 

Doc Al

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Quoting the problem statement in your first post:
The half-life of C-14 is approxiamtely 6000 years.
 
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therefore the answer to c) is 375 years.
 

Doc Al

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joejo said:
therefore the answer to c) is 375 years.
How did you arrive at that answer?
 
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my test is in a few hours...if your planning on helping me out stop confusing me...you could help by telling me what the answer. thanks
 

Doc Al

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If asking how you got your answer confuses you, there's not much I can do.

Try this for problem c. Answer each question in turn:

1) How many half-lives did it take to get from 32g to 2g?

2) How long is each half-life? (that's given)

3) What total time must have passed? (multiply!)
 

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