Question

Egyptian papyrus with 63% of its original carbon-14 atoms
Aboriginal charcoal with 22% of its original carbon-14 atoms.
Mayan headdress with 79% of its original carbon-14 atoms
Neanderthal skull with 3% of its original carbon-14 atoms

What are their ages?

Answer 1

Their ages are:

Egyptian papyrus: 3811.10 years

Aboriginal charcoal: 12510.36 years

Mayan headdress: 1905.55 years

Neanderthal skull: 28997.51 years

Why?

If we need to calculate how old is something, we can use the equation for radioactive decay rate. It's possible using the C-14 half-life (5740 years) as reference.

We must remember, C-14 is used because all the living things take up the element from the atmosphere, when an organism dies, the amount of carbon starts to decay (slowly).

We can calculate the decay rate using the following equation:

Decay rate half life:

$N_{t}=N_{o}*e^{-l*t}$

Where,

$N_{o}$ is the initial number of atoms (undecayed)

$N_{t}$ is the the number of atoms at time (undecayed)

$l$  the decay rate

Now, isolating the decay rate of the formula, we have:

$N_{t}=N_{o}*e^{-l*t}\\\\ln(\frac{N_{t}}{N_{o}})=-l*t$

Also, we can get the value of the decay rate (half life), using the following formula:

$l=\frac{0.693}{5740}=1.207x10^{-4}$

Now, calculating, we have:

Egyptian papyrus: With 63% of its original carbon-14 atoms.

$ln(\frac{63}{100})=-1.207x10^{-4}*t\\\\-0.46=-1.207x10^{-4}*t\\\\t=\frac{-0.46}{-1.207x10^{-4}}=3811.10years$

Aboriginal charcoal: With 22% of its original carbon-14 atoms.

$ln(\frac{22}{100})=-1.207x10^{-4}*t\\\\\\-1.51=-1.207x10^{-4}*t\\\\t=\frac{-1.51}{-1.207x10^{-4}}=12510.36years$

Mayan headdress: With 79% of its original carbon-14 atoms.

$ln(\frac{79}{100})=-1.207x10^{-4}*t\\\\\\-0.23=-1.207x10^{-4}*t\\\\t=\frac{-0.23}{-1.207x10^{-4}}=1905.55years$

Neanderthal skull: With 3% of its original carbon-14 atoms.

$n(\frac{3}{100})=-1.207x10^{-4}*t\\\\\\-3.50=-1.207x10^{-4}*t\\\\t=\frac{-3.50}{-1.207x10^{-4}}=28997.51years$

Have a nice day!

Answer 2

The ages of the samples are obtained from the percentage of radioactive isotopes found after t years.

Given that;

0.693/t1/1 = 2.303/t log No/N

Where;

t1/1 = half life of C-14 = 5670 years

No = original amount of C-14 present

N = amount of C-14 present at time t

t = age of the sample

For Egyptian papyrus, the number of C-14 atoms left after time t is 0.63 No hence;

0.693/5670 = 2.303/t log No/0.63No

1.22 × 10^-4 = 0.462/t

t = 0.462/1.22 × 10^-4

t = 3787 years

The Egyptian papyrus is 3787 years.

For Aboriginal charcoal, the number of C-14 atoms left after time t is 0.22 No hence;

0.693/5670 = 2.303/t log No/0.22No

1.22 × 10^-4 = 1.51/t

t = 1.51/1.22 × 10^-4

t = 12377 years

For Mayan headdress, the number of C-14 atoms left after time t is 0.79 No hence;

0.693/5670 = 2.303/t log No/0.79No

1.22 × 10^-4 = 0.235/t

t = 0.235/1.22 × 10^-4

t = 1926 years

For Neanderthal skull, the number of C-14 atoms left after time t is 0.03 No hence;

0.693/5670 = 2.303/t log No/0.03No

1.22 × 10^-4 = 3.507/t

t = 3.507/1.22 × 10^-4

t = 28746 years

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