1. ## Half life problem

Hello,

I could use some help with the following question.

The half-life of $C^{14}$ is 5730 years. If a sample of $C^{14}$ has a mass of 20 micrograms at time 0, how long will it take until (a) 10 micrograms, and (b) 5 micrograms are left.

Here's what I have:
To solve, I use the exponential decay formula:

$N = N_0e^{kt}$

First, I solve for k:

$10 = 20e^{5730k}$

$\frac{1}{2} = e^{5730k}$

$ln(\frac{1}{2}) = 5730k$

$-ln(2) = 5730k$

$\frac{-ln(2)}{5730} = k$

But, here I am stuck, because the value of
$\frac{-ln(2)}{5730}$ is a very long non-terminating decimal. How am I supposed to use this in the exponential decay formula?

thanks

2. Originally Posted by centenial
Hello,

I could use some help with the following question.

The half-life of $C^{14}$ is 5730 years. If a sample of $C^{14}$ has a mass of 20 micrograms at time 0, how long will it take until (a) 10 micrograms, and (b) 5 micrograms are left.

Here's what I have:
To solve, I use the exponential decay formula:

$N = N_0e^{kt}$

First, I solve for k:

$10 = 20e^{5730k}$

$\frac{1}{2} = e^{5730k}$

$ln(\frac{1}{2}) = 5730k$

$-ln(2) = 5730k$

$\frac{-ln(2)}{5730} = k$

But, here I am stuck, because the value of
$\frac{-ln(2)}{5730}$ is a very long non-terminating decimal. How am I supposed to use this in the exponential decay formula?

thanks
Saying the half life is "5730 years" is only an approximation to 3 significant figures. Cut your "very long non-terminating decimal" to 3 (or 4 if you want to be very careful) significant figures.

3. Originally Posted by HallsofIvy
Saying the half life is "5730 years" is only an approximation to 3 significant figures. Cut your "very long non-terminating decimal" to 3 (or 4 if you want to be very careful) significant figures.
Thanks for your reply. The value of $-ln(2)/5730$ is "-0.000120968094338559390823251679137552629681588156 083814180..."

So, I would cut it down to "0.0001"?

4. This problem is a lot easier than you are making it out to be! You just need to understand what the definition of "half life" is - then the answers become obvious. The half-life is the time it takes for the concentration of particles to be reduced by half from the original. If you start with 20 micrograms of C14, and its half-life is 5730 years, that means that after 5730 years you will have half or 20 grams, or 10 micrograms of C14. If you wait another 5730 years after that, the 10 micrograms is reduced to 5. So the answers are obvious.

If you want to see how the math works - you need to use the correct formula. The correct formula is:

$
N = N_0 (\frac 1 2 ) ^{t/k}
$

where k is the half life. So for the first question:
$
10 = 20 (\frac 1 2 ) ^{t/5730}
$

$
\frac 1 2 = (\frac 1 2 )^{t/5730}
$

t = 5730.

5. Originally Posted by centenial
Thanks for your reply. The value of $-ln(2)/5730$ is "-0.000120968094338559390823251679137552629681588156 083814180..."

So, I would cut it down to "0.0001"?
No, I said "3 or 4 significant figures", not 4 decimal places. To three significant figures would be $1.21 x 10^{-4}= .000121$ and to four significant figures it would bbe $1.210 x 10^{-4}= .0001210$.

Are you using a calculator to do this? Most calculators today have the ability to store numbers and it would be better to keep the number in calculator memory rather than writing it down and keeping all decimal places.

6. Originally Posted by HallsofIvy
No, I said "3 or 4 significant figures", not 4 decimal places. To three significant figures would be $1.21 x 10^{-4}= .000121$ and to four significant figures it would bbe $1.210 x 10^{-4}= .0001210$.

Are you using a calculator to do this? Most calculators today have the ability to store numbers and it would be better to keep the number in calculator memory rather than writing it down and keeping all decimal places.
Just be aware that while this is good advice on the use of significant digits, it's still the wrong answer.