Question

A fabric remnant from a burial site has a 14c/12c ratio of 0.662 of the original value. how old is the fabric?

Answer 1

The half-life of carbon-14 is 5730 years, so you have
.. 0.662 = 2^(-t/5730) . . . . . . . . . . . . . . . . . t is age in years (before present)
.. log(0.662) = -t/5730*log(2)
.. t = -5730*log(0.662)/log(2) ≈ 3410 . . . . years

Answer 2

The age of the fabric is $\boxed{3.4 \times {{10}^3}{\text{ years}}}$.

Further  explanation:

Radioactive decay:

Also known as nuclear decay, radioactivity, nuclear or radioactive disintegration. It is the process due to which an unstable atomic nucleus releases its energy in the form of various particles such as alpha particles, beta particles, and gamma particles.

Half-life is defined as the time after which the amount of radioactive substance becomes half of its initial value. It is represented by ${t_{1/2}}$.

The relationship between decay constant and the half-life is of carbon-14 (14 C) is as follows:

$k = \dfrac{{{\text{ln2}}}}{{{t_{1/2}}}}$          …… (1)

Here,

k is the decay constant of 14 C.

${t_{1/2}}$ is the half-life of 14 C.

                                                     

The value of half-life of carbon-14 is 57330 years.

Substitute 5730 yr for ${t_{1/2}}$ in equation (1).

$\begin{aligned}k&= \frac{{{\text{ln 2}}}}{{5730{\text{ yr}}}}\\&= \frac{{0.69}}{{5730{\text{ yr}}}}\\&= 0.000120968\;{\text{y}}{{\text{r}}^{ - 1}}\\\end{aligned}$

The relationship between decay constant and time for carbon-14 is as follows:

$k = \dfrac{1}{t}\ln \left( {\frac{{{N_0}}}{{{N_t}}}} \right)$    ......(2)

Here,

k is the decay constant of carbon-14.

t is the time or age of carbon-14.

${N_0}$ is the initial amount of carbon-14.

${N_t}$ is the final amount of carbon-14.

Rearrange equation (2) to calculate time.

$t =\dfrac{1}{k}\ln \left( {\frac{{{N_0}}}{{{N_t}}}} \right)$       ......(3)

But the value of $\left( {\dfrac{{{N_0}}}{{{N_t}}}} \right)$ is the reciprocal of the ratio of 14 C to 12 C. So its value can be calculated as follows:

$\begin{aligned}\left( {\frac{{{N_0}}}{{{N_t}}}} \right) &= \frac{1}{{0.662}}\\&= 1.5105\\\end{aligned}$

The value of k is $0.000120968\;{\text{y}}{{\text{r}}^{ - 1}}$.

The value of $\left( {\dfrac{{{N_0}}}{{{N_t}}}} \right)$ is 1.5105.

Substitute these values in equation (3).

$\begin{aligned}t&= \left( {\frac{1}{{0.000120968\;{\text{y}}{{\text{r}}^{ - 1}}}}} \right)\ln \left( {1.5105} \right)\\&= \left( {\frac{1}{{0.000120968\;{\text{y}}{{\text{r}}^{ - 1}}}}} \right)\left( {0.4124} \right)\\&= 3.4 \times {10^3}{\text{ years}}\\\end{aligned}$

Therefore the age of the fabric is $3.4 \times {10^3}{\text{ years}}$.

Learn more:

1. What nuclide will be produced in the given reaction? brainly.com/question/3433940

2. Calculate the nuclear binding energy: brainly.com/question/5822604

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Nuclear chemistry

Keywords: half-life, decay constant, fabric, age, 3.4*10^3 years, k, t, t1/2, N0, Nt, 1.5104, 0.662