Question

The half-life of Carbon-14 is 5730 years. Suppose a fossil is found with 50% of Carbon-14 as compared to a living sample. How old is the fossil?
a) 572 years
b) 57 years
c) 5730 years
d) 5728 years

Answer 1

The answer is C. Because it takes 5730 years for half of a sample of carbon-14 atoms to decay. It says that 50% of the carbon atoms have decayed so that means that 5730 years have elapsed for that fossil.

I hope that helps.

Answer 2

The fossil is 572 years old which is option A

Data Given;

• half life (T1/2) = 5730 years
• The ratio between a fossil to a living sample = 50% = 0.5
• t = ?

Disintegration Constant

To solve this question, let's find the disintegration constant.

$T_\frac{1}{2} = ln2/ y\\ y = in2 / T_\frac{1}{2}\\ y = 0.693/5730\\ y = 0.00012$

Let's plug this into the radio-decay formula

$[\frac{N}{N_o}]= e^-^y^t\\ (0.5)=e^-^0^.^0^0^0^1^2^*^t$

Let's take the natural log of both sides

$ln(0.5)=(0.00012)t\\\ -0.693=-0.00012t\\ t=572$

The fossil is 572 years old.

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