Date An Object Using Carbon Dating, Beta Decay

[breakingaway]

Homework Statement

An old wooden tool, containing 75.0 grams of carbon, is found in an ancient tomb. The tool emits 500 electrons/minute from the beta decay of ¹⁴₆C. How old is the wood from which the tool was constructed? Given: The half-life of ¹⁴₆C is 5730 years, the ratio of ¹⁴₆C to ¹²₆C in living plants is 1.30 x 10^-12, 1 year = 3.15576 x 10⁷s, NA= 6.0221415 x 10²³atoms/gram-mole.

Homework Equations

R = $\lambda$N = $\lambda$N₀e^-$\lambda$t = R₀e^-$\lambda$t
R = rate of decay, $\lambda$ = decay constant, N₀ = # of radioactive nuclei at t=0, N = number of radioactive nuclei now, R₀ = decay rate at t=0

T_1/2 = ln|2|/$\lambda$
T_1/2 = half life

# of atoms of ^A_ZX = N = m * N_A / A
m = mass of sample, A = atomic mass of element, Z is atmoic # of element, N = number of atoms, N_A = Avogadro's Constant

The Attempt at a Solution

R = 500
N₀ = 75 * 1.3*10^-12 * N_A / 14 = 4.194*10¹²
T_1/2 = ln|2|/$\lambda$ - > $\lambda$= ln|2|/T_1/2 = 3.83*10^-12
R = 500 = $\lambda$N --> N = 1.305*10¹⁴
R = $\lambda$N₀e^-$\lambda$t --> Solve for t
I get t=3.0863*10⁴years

The correct answer is: 6.7*10³ years

 

[TSny]

N₀ = 75 * 1.3*10^-13 * N_A / 14 = 4.194*10¹²

If the ratio 1.3*10^-12 is the ratio of the number of C14 to C12, then you would first want to find the number of C12 and then multiply by 1.3*10^-12.

R = 500 = $\lambda$N --> N = 1.305*10¹⁴

It's a good idea to carry along your units in the calculation. I think you'll see that the units are mismatched here.

 

[breakingaway]

If the ratio 1.3*10^-12 is the ratio of the number of C14 to C12, then you would first want to find the number of C12 and then multiply by 1.3*10^-12.


It's a good idea to carry along your units in the calculation. I think you'll see that the units are mismatched here.

The 10^-13 was a typo. I'm still having issues with this problem. I tried it again doing this:

N₀ = 75g * 1.3*10^-12 * (6.022*10²³)gram-mol / (14gram-mol) = 4,19 * 10¹²

T_1/2 = ln|2| / $\lambda$, solving for $\lambda$ gives $\lambda$ = 3.833*10^-12

R=$\lambda$N₀e^-$\lambda$t solving for t.
Plugging in R as 500minutes / (60seconds/minute) = 8.3333 decays/second.
This is giving me 7.1455 * 10³ years.

I think my mistake might be that 500 electrons/minute doesn't equal 500 decays/minute?

 

[TSny]

I think my mistake might be that 500 electrons/minute doesn't equal 500 decays/minute?

I think that's ok.

Try calculating N_o as

N₀ = 75g * 1.3*10^-12 * (6.022*10²³)gram-mol / (12gram-mol)

Think about why you would want to calculate it this way.

 

[breakingaway]

I think that's ok.

Try calculating N_o as

N₀ = 75g * 1.3*10^-12 * (6.022*10²³)gram-mol / (12gram-mol)

Think about why you would want to calculate it this way.

I did this, and got the correct answer, but I am still somewhat confused as to why.

My thought was that the C14 is the radioactive isotope, I should be trying to determine how much C14 is currently still present. To find that I need to figure out how much has decayed into C12. So I know the ratio at t=0, which let's me solve for N₀. Why would I solve for N₀ as C12 instead of C14 if C14 is the radioactive one?

Also, a huge thank you for the help!

 

[TSny]

(75g / 12g mol^-1) * (6.022*10²³ atoms mol^-1) gives the number of C12 atoms in the 75 g sample. You then multiply this by 1.3*10^-12 to get the number of C14 atoms that were originally present in the 75 g sample.

 

[breakingaway]

(75g / 12g mol^-1) * (6.022*10²³ atoms mol^-1) gives the number of C12 atoms in the 75 g sample. You then multiply this by 1.3*10^-12 to get the number of C14 atoms that were originally present in the 75 g sample.

Ah, OK that makes sense now. Thank you. I'm guessing that since there are so few C14 atoms, its fair to make the assumption that the entire 75g sample is C12 when finding the number of C12 atoms?

 

[TSny]

Ah, OK that makes sense now. Thank you. I'm guessing that since there are so few C14 atoms, its fair to make the assumption that the entire 75g sample is C12 when finding the number of C12 atoms?

Yes, that's right. Good.