Question 5d643

Jan 18, 2017

4052 years old.

Explanation:

${2}^{.5}$ = 1.414

25% is half of half life. so 2 to the .5 of a half life.

So the amount of the carbon left. is $\frac{1}{1} , 414 \times 5730$ = 4052 years.

Feb 6, 2017

The sample is 2380 years old.

Explanation:

We can express the rate law for a first-order decay as

${N}_{t} / {N}_{0} = {\left(\frac{1}{2}\right)}^{n}$

where

${N}_{t}$ is the amount remaining after time $t$.
${N}_{0}$ is the amount present at the start ($t = 0$).
$n$ is the number of half-lives (t_½)

We can rewrite this as

${N}_{0} / {N}_{t} = {2}^{n}$

If the concentration of carbon-14 has decreased by 25 %, then 75 % remains.

$\frac{100}{75} = {2}^{n}$

$\ln \left(\frac{100}{75}\right) = n \ln 2$

$n = \frac{0.2877}{\ln} 2 = 0.4150$

The sample has decayed for 0.4150 half-lives.

0.4150t_½ = "0.4150 × 5730 yr" = "2380 yr"#