The half-life of carbon-14 is 5730 years. How long does it take for 3.6 grams of carbon-14 to be reduced to 0?

Question

Mason Allen · Accepted Answer

It takes an indefinitely long time.

For every half-life that passes, the mass of the carbon-14 is reduced by half. Suppose there were #x# amount of C-14 initially. After 1 half-life, the amount of C-14 left is #x/2#. After 2 half-life, the amount of C-14 left is #x/4#. After 3 half-life, the amount of C-14 left is #x/8#. After #n# half-life, the amount of C-14 left is #x/(2^n)#. Notice that the amount of C-14 left (#x"/"2^n#) can be brought as close as we like to zero, by letting #n# be a sufficiently large number. However, it will never reach zero no matter what #n# we use.

Aaron Anderson · Answer

undefined To calculate the time it takes for a certain amount of carbon-14 to decay to zero, you can use the formula for exponential decay:

$$N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{\frac{1}{2}}}}$$

Where:
- $N(t)$ is the amount of carbon-14 remaining after time $t$
- $N_0$ is the initial amount of carbon-14
- $t$ is the time elapsed
- $T_{\frac{1}{2}}$ is the half-life of carbon-14

In this case, $N(t)$ is 0, $N_0$ is 3.6 grams, and $T_{\frac{1}{2}}$ is 5730 years.

Plugging these values into the formula:

$$0 = 3.6 \times \left(\frac{1}{2}\right)^{\frac{t}{5730}}$$

$$0 = \left(\frac{1}{2}\right)^{\frac{t}{5730}}$$

Since any non-zero number raised to the power of 0 is 1, we can rewrite the equation as:

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

Taking the natural logarithm of both sides to solve for $t$:

$$\ln(1) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)$$

$$0 = \frac{t}{5730} \times \ln\left(\frac{1}{2}\right)$$

$$0 = \frac{t}{5730} \times -0.693$$

$$0 = t \times -0.693$$

$$0 = t$$

So, it takes 0 years for 3.6 grams of carbon-14 to decay to 0.