# Homework Help: Carbon dating

1. Jan 6, 2014

### QuarksAbove

1. The problem statement, all variables and given/known data
Suppose you have a 1.4g sample of old charcoal. It produces 0.7 beta decays per minute. How old is the charcoal.

Given:
1g of carbon current day has 6.36x1010 atoms of 14C

2. Relevant equations

N = Noe-rt

N = number of atoms in the sample (current-day)
No = original number of atoms (i.e. at time of death)
r = decay rate = 1.21x10-4
t = time

3. The attempt at a solution

I know that No = 1.4*6.36x1010 = 8.904x1010
r = 1.21x10-4

what I don't understand is what to do with the 0.7 decays per minute. I know I need to solve for N before I can solve for t, but I'm stuck. As soon as I solve for N, it's plug and chug.

Any help would be much appreciated!

2. Jan 6, 2014

### SteamKing

Staff Emeritus
It would be helpful to know the units of the decay rate r.

3. Jan 6, 2014

### ehild

If you have N C14 atoms, how many of them decays in unit time? How is it related to the decay rate?

ehild

4. Jan 6, 2014

### QuarksAbove

r = 1.21x10-4 /year

If I have N C-14 atoms, then 0.7 of those N atoms decay each minute.. 0.7 decay/minute = 367920 atom decays per year.

I'm not sure how it's related to the decay rate. Sorry, I haven't done a problem with two rates before and it's really confusing.

5. Jan 6, 2014

### ehild

N is the number of the 14C atoms in the sample at present. N0 was the number of atoms when the sample got isolated. N=N0e-rt, so dN/dt=-rN0e-rt=-rN.

rN atoms decays in a year, that is about 368000 in the 1.4 g sample. You know r. What is N then?
Originally there were 6.36x1010atoms of 14C in 1 g sample. Current-day means "fresh" sample, which can interact with the surroundings, so has supply of 14C. The charcoal is isolated, so the number of 14C atom decreases with time.

ehild