1. Apr 2, 2005

### Naeem

Q. The half life of carbon-14 is 5730 years. If the original amount of carbon-14 in a particular living organism is 20 grams and that found in a fossil is 6 grams. determine the approximate age of the fossil.

Here is what I did:

A(t) = A0 e^kt

1/2 A0 = A0 e^5730k

Find k by taking ln of both sides

A0 = 20 ( for living organism )

A0 = 6 ( for fossil )

Using A0 = 6 can we find 't' the age of the fossil.

Is this idea correct? Why A0 = 20 is given in the problem?

Any ideas:

2. Apr 2, 2005

### dextercioby

That exponent must be negative,since the C-14 quantity is diminishing with time...

$$A(t)=A_{0}e^{-kt}$$

U need to find "t",knowing the halflife (that should give you "k").

Daniel.

3. Apr 2, 2005

### saltydog

The formula for radioactive decay is:

$$\frac{dA}{dt}=-kA$$

You know, the rate decreases proportional to the amount present. Solving this we get:

$$A(t)=A_0e^{-kt}$$

Now, the half-life is related to the decay constant k as follows:

$$t_{1/2}=\frac{ln(2)}{k}$$

So, you can plug this into the equation, supply the amounts for the half-life and the initial amount and what's left, do some algebra, you can do that, and then solve for the time when only 6 grams are left.

Did I give that one away? You know what, you need to figure out what the decay constant is. Ignore that man behind the curtain . . . do what Daniel said.

Last edited: Apr 2, 2005
4. Apr 2, 2005

### Giuseppe

Hmm

Technically speaking, isn't that formula just for 1st order reactions. When you start dealing with zero order and second order reactions, concentrations are involved and make a difference.

5. Apr 2, 2005

### saltydog

Please Giuseppi, elaborate with an example if you might.

6. Apr 2, 2005

### Gokul43201

Staff Emeritus
Generally speaking, that formula is for a first order reaction. However, there is absolutely no problem with this, since readioactive decay IS AN EXAMPLE OF A FIRST ORDER REACTION.

An n'th order reaction in a single reactant is defined by the rate equation :

$$\frac {dX}{dt} = -kX^n$$

Last edited: Apr 2, 2005
7. Apr 2, 2005

### HallsofIvy

The way I would do this problem is: If the half life is 5730 years, then the amount left, of C initially, after t years, is $$A(t)= C\left( \frac{1}{2}\right)^{\frac{t}{5730}}$$. (Notice that t/5730 just measures the number of times the quantity is "halved".)

You are told that C= 20 and that A(t)= 6. Solve $$20\left(\frac{1}{2}\right)^{\frac{t}{5730}}= 6$$ for t.

Divide both sides by 20 and then take a logarithm of both sides.

8. Apr 3, 2005

### Naeem

Yes, I did the above, and

ln ( 6/20) = t/5730 . ln (1/2)

-1.2039 = (-0.693) ( t/5730)

-1.2039 =-0.0001t

t = -1.2039/-0.001
= 12039 years, is this correct.

9. Apr 3, 2005

### saltydog

Naeem, let's think about this for a while: The half-life is 5730 and you start with 20 grams. So after 5730 years you'll have 10 grams, and after 11460 years you'll have 5 grams. So at 12039 years you'll have less than 5 grams.

10. Apr 3, 2005

### Naeem

I get what you are saying,

I believe what I have done is right.

11. Apr 3, 2005

### saltydog

Naeem, I think you should use more decimal digits to get a more accurate answer.

12. Apr 4, 2005

### dextercioby

There's something devious with his answer.It should be ~9953 yrs...

Daniel.

13. Apr 4, 2005

### Gokul43201

Staff Emeritus
No, it's not.

In step 3, you write : 0.693/5730 = 0.0001

You make an approximation to one significant figure !!!! This can give you an error as big as 100% in the final number (in this case, you have an error of about 20%)

Dex, what's devious ? I get the same number you get.

14. Apr 4, 2005

### dextercioby

I meant his answer,not mine (=yours)...

Daniel.

15. Apr 4, 2005

### Naeem

Yeah, I know what Gokul said.

Now I too get 9953 years.

BTW, what actually does devious mean. Don't have a dictionary handy, but doesn't sound good.

May be it might be close to deceiving or something,...... don't know

Thanks, for your help on this, you all !!!

16. Apr 4, 2005

### dextercioby

Daniel.

17. Apr 6, 2005

### iasever

HELP!! with this SILLY looking RADIOACTIVITY problem

I was presented with a radioactivity problem which does not make any sense to me. Hope it does to someone in here:

A radioactive element releasing 8 rays (whatever this means) is giving a Geiger count of 188 per minute at the beginning of a test. 40 minutes down the line the count drops to 53. What is the half life??? (The answer is apparently 20 but HOW?)

Thanks for any help

18. Apr 6, 2005

### dextercioby

The way i see it...It's more like

$$53=188\left(\frac{1}{2}\right)^{\frac{40}{\tau_{\frac{1}{2}}}}$$

I dunno where that 8 comes into turn.I think they might mean counting the rays.Which initially (in the first minute) would be 188 and after 40mins it would drop to 53,which is just what my eq.stated...

Daniel.

19. Apr 7, 2005

### iasever

Yes but the result of the equation is not 20 minutes. I had used the same equation you quoted and that gives you an answer close to 22 minutes, not 20!

20. Apr 7, 2005

### dextercioby

$$\allowbreak x=21.\,89778\,94746\,45439\,18763\,33516\,196$$ that's the answer with 30 sig.digits.

I dunno how they got the 20...

Daniel.