Radiocarbon Dating: Determining Sample Age of 41000-Yrs

[blue_lilly]

Homework Statement

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 6 12 C atoms remains?

Homework Equations

(N1/No)=e^-λt

The Attempt at a Solution

Variables:
t = 41,000 yrs
T1/2 of Carbon = 5730 yrs
ln2 = .693

I basically plugged in the numbers and solved because you are given all the variables.

(N1/No)=e^-λt
(N1/No) = e^-(ln2/T1/2)t
(N1/No) = e^-(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(1.2094E-4)(41,000 yrs)
ln(N1/No) = -4.95863
(N1/No) = e^(-4.95863)
(N1/No) = .0102548​

I then calculated for the % because the answer .012548 is a fraction.

.0102548/100 %
=1.025%​

However that answer is incorrect and I'm not exactly sure why. I made sure that I was using the (ln)-function instead of the (log)-function. I don't know if it is my math or if I'm putting it into the website incorrectly.

Any help would be greatly appreciated!

 

[SammyS]

Homework Statement

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 6 12 C atoms remains?

Homework Equations

(N1/No)=e^-λt

The Attempt at a Solution

Variables:
t = 41,000 yrs
T1/2 of Carbon = 5730 yrs
ln2 = .693

I basically plugged in the numbers and solved because you are given all the variables.

(N1/No)=e^-λt
(N1/No) = e^-(ln2/T1/2)t
(N1/No) = e^-(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(1.2094E-4)(41,000 yrs)
ln(N1/No) = -4.95863
(N1/No) = e^(-4.95863)
(N1/No) = .0102548​

I then calculated for the % because the answer .012548 is a fraction.

.0102548/100 %
=1.025%​

However that answer is incorrect and I'm not exactly sure why. I made sure that I was using the (ln)-function instead of the (log)-function. I don't know if it is my math or if I'm putting it into the website incorrectly.

Any help would be greatly appreciated!

I get that e^(-4.95863) ≈ 0.00702 .

 

[blue_lilly]

I get that e^(-4.95863) ≈ 0.00702 .

You are correct, I must have copied it down wrong.

So then, I would take .00702 and divide by 100 to get the Percentage.

.00702/100 = 7.0225E-5%

This is correct, right?

 

[SammyS]

You are correct, I must have copied it down wrong.

So then, I would take .00702 and divide by 100 to get the Percentage.

.00702/100 = 7.0225E-5%

This is correct, right?

Isn't 1/10 equal to 10 % ?

You don't get that by dividing by 100 .

 

[blue_lilly]

Isn't 1/10 equal to 10 % ?

You don't get that by dividing by 100 .

Where are you getting the 10% from?

I was dividing by 100 because initially you start out with 100% of the material. Or am I incorrect with that idea?

 

[haruspex]

Where are you getting the 10% from?

I was dividing by 100 because initially you start out with 100% of the material. Or am I incorrect with that idea?

SammyS was illustrating that diving by 100 does not give the percentage.
What is 0.1 as a percentage? How is it calculated?