Carbon-14 has a half-life of 5770 years. If a fossil is 23,080 years old and it has 3kg of Carbon-14, how much carbon-14 did it originally have?

Question

Naomi Aronson · Accepted Answer

#"48 kg"#

The nuclear half-life of a radioactive isotope indicates how long it will take for a sample of the isotope to shrink to half its original size.

In essence, any sample you start with will be halved with every passing of a half-life. So if you start with a mass #A_0# of your isotope, you can say that you'll be left with

#A_0 * 1/2 = A_0/2 -># after one half-life

#A_0/2 * 1/2 = A_0/4 -># after two half-lives

#A_0/4 * 1/2 = A_0/8 -># after three half-lives

#vdots#

and so forth. Consequently, you can determine a correlation between the number of half-lives that have elapsed in a specific amount of time and the amount of your initial sample that remains undated.

#color(blue)(A = A_0 * 1/2^n)" "#, where

#A# - the amount left undecayed #n# - the number of half-lives that have passed

In your case, you know that the fossil is #"23,080"# years old and that carbon-14, the isotope of interest, has half-life of #"5,770"# years.

This implies that you calculate the number of half-lives that have elapsed during this duration.

#n = (23080 color(red)(cancel(color(black)("years"))))/(5770color(red)(cancel(color(black)("years")))) = 4#

So, you know that #"3.0 kg"# of carbon-14 are left undecayed after the passing of four half-lives, which means that the sample originally contained

#A = A_0 * 1/2^n implies A_0 = A * 2^n#

#A_0 = "3.0 kg" * 2^4 = color(green)("48 kg")#

Noah Aldrich · Answer

undefined To find the original amount of Carbon-14 in the fossil, we can use the formula for exponential decay: $ N(t) = N_0 \times (1/2)^{t/t_{\text{half-life}}} $, where $ N(t) $ is the amount of Carbon-14 present at time $ t $, $ N_0 $ is the initial amount of Carbon-14, $ t $ is the time elapsed, and $ t_{\text{half-life}} $ is the half-life of Carbon-14.

Given:
- $ t_{\text{half-life}} = 5770 $ years
- $ t = 23080 $ years
- $ N(t) = 3 $ kg

Using the formula, we can solve for $ N_0 $:

$$ 3 = N_0 \times (1/2)^{23080/5770} $$

$$ 3 = N_0 \times (1/2)^{4} $$

$$ 3 = N_0 \times (1/16) $$

$$ N_0 = 3 \times 16 $$

$$ N_0 = 48 $$

The fossil originally had 48 kg of Carbon-14.

Serenity Allen · Answer

undefined To determine how much carbon-14 the fossil originally had, we can use the concept of exponential decay and the formula for calculating the amount of a substance remaining after a certain amount of time has passed.

Given that carbon-14 has a half-life of 5770 years, we can use the formula:

$$ N = N_0 \times (0.5)^{\frac{t}{t_{\frac{1}{2}}}} $$

Where:
- $ N $ is the amount of substance remaining after time $ t $.
- $ N_0 $ is the initial amount of the substance.
- $ t $ is the elapsed time.
- $ t_{\frac{1}{2}} $ is the half-life of the substance.

Substituting the given values:
- $ N = 3 $ kg (amount remaining after 23,080 years)
- $ t = 23,080 $ years
- $ t_{\frac{1}{2}} = 5770 $ years

We can solve for $ N_0 $, the initial amount of carbon-14:

$$ 3 = N_0 \times (0.5)^{\frac{23080}{5770}} $$

$$ 3 = N_0 \times (0.5)^4 $$

$$ 3 = N_0 \times (0.0625) $$

$$ N_0 = \frac{3}{0.0625} $$

$$ N_0 = 48 $$ kg

Therefore, the fossil originally had 48 kg of carbon-14.