Question

Hello pls help!!

The half-life of carbon-14 is 5,730 years. Suppose a fossil is found with 20 percent as much of its carbon-14 as compared to a living sample. How old is the fossil?
A. 13,307 years
B. 3,235 years
C. 1,331 years
D. 32, 346 years

Please help it's very urgent!!

Answer 1

13,307 years

Explanation:

The expression for the decay of a radiactive material is:

A = B*(1/2)^(t/HL),

where A is the amount remaining, B is the initial amount, t is time (in years), and HL is the half life (in years).

We learn that A is 20% of B, so let's rewrite for that:

0.20B = B*(1/2)^(t/HL)

and we can divide both sides by B to leave us:

0.20 = (1/2)^(t/HL)

The HL is 5730 years:

0.20 = (1/2)^(t/5730)

We need to solve this expression for t, the time (in years) that is required before we have 20% of the carbon remaining.

The answer I get, by graphing, is 13,305 years, The closest to this is A) 13,307 years. An algebraic solution might result in a slightly different number.