Chapter 21, Problem 21.82QA

Interpretation Introduction

To find:

a)  The percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE.

b) The ratio of carbon-14: carbon-12 in the blackened grains compared with that of the grain harvested last year.

Answer to Problem 21.82QA

Solution:

a)  The percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE = 3%

b)  The ratio of carbon-14: carbon-12 in the blackened grains from the Jericho site (1315 $\pm 13$ years) compared with that of the grain harvested last year is 1: 0.27

Explanation of Solution

1) Concept:

From the half-life of nuclear decay, and the ratio of isotopes, we can calculate the isotopic ratios backwards in time. This is a powerful tool for radioactive dating techniques. This technique is used extensively for geological dating of various fossils, and rocks.

Ratio of C-14 to C-12 is a powerful tool used for radioactive dating..

2) Formula:

i) $t_{\frac{1}{2}}=\frac{0.693}{k}$

ii) $ln\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=-kt$

where, ${\left[X\right]}_t$ is the final concentration, ${\left[X\right]}_0$ is initial concentration, $k$ is rate constant,  $t$ is time, and $t_{\frac{1}{2}}$ is half-life

3) Given:

i) $t1=1360 years, t2=1628 years$

ii) $t=1315\pm 13 years$      

4) Calculations:

The half-life of $C$ is taken from the Table 21.2 as $5730 years$.

So, the rate constant is calculated as follows:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$5730 yr=\frac{0.693}{k}$

$k=\frac{0.693}{5730 yr}=1.209\times {10}^{-4}{yr}^{-1}$

Part a)

Calculation using $t1=1360 years$:

 $ln\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=-kt$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=e^{-kt}=e^{\left(-1.209\times {10}^{-4}{yr}^{-1}\times 1360 yr\right)}=0.84$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=85\%$

Calculation using $t2=1628 years:$

 $ln\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=-kt$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=e^{-kt}=e^{\left(-1.209\times {10}^{-4}{yr}^{-1}\times 1628 yr\right)}=0.82$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=82\%$

Therefore, the percent difference is $85\%-82\%=3\%$

Thus the percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE = 3%

Part b)

$t=1315\pm 13 years$

So, $t1=1315-13=1302 years$ and $t2=1315+13=1328 years$

Calculation using $t1=1302 years$:

 $ln\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=-kt$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=e^{-kt}=e^{\left(-1.209\times {10}^{-4}{yr}^{-1}\times 1302 yr\right)}=0.8543$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=85.43\%$

Calculation using $t2=1328 years:$

 $ln\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=-kt$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=e^{-kt}=e^{\left(-1.209\times {10}^{-4}{yr}^{-1}\times 1328 yr\right)}=0.8516$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=85.16\%$

Therefore, the percent difference is $85.43\%-85.16\%=0.27\%$

For grain harvested last year, take t = 1 year.

k = $1.209\times {10}^{-4}{yr}^{-1}$

$\frac{{\left[X\right]}_t}{{\left[X\right]}_0}=e^{-kt}$= 0.999

Thus the ratio of carbon-14: carbon-12 in the blackened grains from the Jericho site (1315 $\pm 13$ years) compared with that of the grain harvested last year is 1: 0.27

Conclusion:

The ratio of carbon-14: carbon-12 is calculated using the first order radioactive decay equation.