Discussion in 'Physics' started by srinipadma45, Apr 17, 2008.

Mar 7, 2008
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1) An old wooden tool is found to contain only 11.9 percent of 14C that a sample of fresh wood would. How many years old is the tool?
2) A specimen taken from the wrappings of a mummy contains 7.02 g of carbon and has an activity of 1.34 Bq. How old is the mummy? Determine its age in years assuming that in living trees the ratio of 14C/12C atoms is
1.23E-12.

2. ### Mark44 Well-Known Member

Nov 26, 2007
626
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These sound like homework problems. If so, look at your notes and/or text to see if there are any similar example problems. Show us how you think you should do these problems and we'll be happy to help you out, but we are not in the business of doing your work for you.

Mar 7, 2008
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1)
For 14 C T 1/2 = 5570 years N/N0 = 11.9/100 = 0.119

N / N0 = e- λ t 2.303 x In 0.119 = -λt λ =0.6931 / 5570 =1.244 x 10 -4

t = (2.303 x In0.119 x 5570) / 0.6931 = 17110 years

2)
R = 1.34 Bq = 1.34 decays / second Mass of sample = 7.02 g

For 14 C T 1/2 = 5700 years (approximate value)

= 5700 x 365 x 24 x 60 x 60 =1.8 x 10 11 seconds

T 1/2 = 0.6931 / λ

λ = 0.6931 / 1.8 x 10 11 = 3.86 x 10 -12 per second

Number of atoms contained in 7.02 g of carbon (12 C atoms)

= (7.02 x 6.023 x 10 23) /14 = 3.02 x 10 23

14 C atoms / 12 C atoms = 1.23 x 10-12

No of 14 C atoms (radioactive) = No of 12 C atoms x 1.23 x 10-12

N0 = 3.02 x 10 23 x 1.23 x 10-12 = 3.7146 x 10 11

R = Nλ

N = R/λ = 1.34 / 3.86 x 10 -12 = 3.4715 x 10 11

t = 3.323 x 5700 x In (3.4715 x 10 11 / 3.7146 x 10 11 ) = 557 years

4. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Let's take a look at problem 1 first. How did you get from this:
N / N0 = e- λ t

to this:
2.303 x In 0.119 = -λt ?

It looks like you took the natural log of each side (BTW, it is ln, not In), so you should have ln(N/N0) = -λt

Where did 2.303 come from?

Your work would be easier to read if you used something to show when you're working with exponents. So instead of e - λt, which looks like you're subtracting λt from e, you could write e^(-λt). And instead of 1.244 x 10 -4, you could write 1.244 x 10^(-4).

Mark

5. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Everything seems reasonable until the last line above. Where does 3.323 come from? I don't see it anywhere in your previous work.

In that last line, are you using t = [ln(N/N0)]/(-λ) ?

Also, please clarify for me what N and N0 represent. I think that N0 = the number of 12-C atoms when the linen was made, and N = the number of 12-C atoms now. Is that how you're using them?

Mark

6. ### Mark44 Well-Known Member

Nov 26, 2007
626
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If this is an Egyptian mummy, your answer of 570 years is too small. An age of 3000 to 4000 years would be more reasonable.

On this other hand, nothing in the problems says that the mummy is Egyptian. There have been mummies found in the Andes for which this age is reasonable.

Mar 7, 2008
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For converting from natural log to log to the base 10 ,it has to be multiplied by a value 2.303 and 2.303/0.6931 gives 3.323

8. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
It would have been helpful to add what you were doing to convert from natural logs to log base-10.

At any rate, I don't see anything obviously wrong with your work, although there are a lot of calculations, so I might have missed something.

Thanks for putting in the effort to show me what you did!
Mark

Mar 7, 2008
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But my answer for the 2nd problem is incorrect. Can anybody help?

10. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Are you off by a little bit or by a lot?
I get a slightly different answer than yours, 549 years, as opposed to your answer, 557 years. I got this result by not rounding until the last step.

I've checked all of your calculations, and we both get the same result (with small rounding differences). If you haven't done so already, make sure that the numbers you're working with are the same as were given in the problem.

Finally, not all answers in the back of the book are correct (I'm assuming that's where you're seeing the "correct" answer). I taught college mathematics for 19 years, and have seen more than a few errors and typos in the answers.
Mark