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?

Answer 1

Answer:

5732 years

Explanation:

Radioactive decay can be determined by using the equation:

$N_t = N_0e^{- \lambda t}$

where;

$N_t$ = number of decayed atoms at time (t)

$N_0$ = initial number of decayed atoms

$\lambda$ = decay constant

So, if we equate the natural log of the above; we have:

$In(N_t) = In(N_0)-\lambda t$

$\frac{In(N_t)} { In(N_0)}} = -\lambda t$

where;

$\lambda$ = $\frac{0.693}{t_{1/2}}$

$\lambda$ = $\frac{0.693}{5730}$

$\lambda$ = $1.209*10^{-4$

$In(\frac{50}{100}) =-(1.209 *10^{-4})*t$

$-0.693 = -(1.209*10^{-4})*t$

$t= \frac{0.693}{1.209*10^{-4}}$

t = 5732.01 years

t = 5732 years.

Hence, the fossil is 5732 years old.