Chapter 42: Q.47b. (page 1237)

What is the age in years of a bone in which the $^{14} C^{12} C$ ratio is measured to be $1.65 \times 10^{- 13}$ ?

Short Answer

Expert verified

The age of bone is 17,058.783 years.

Step by step solution

Step.1.

Ratio of $^{14} C$ to $^{12} C$ in living organisms is,

$\frac{N_{0} (^{14} C)}{N_{0} (^{12} C)} = 1.3 \times 10^{- 12}$

Ratio of $^{14} C$ to $^{12} C$ in bone is,

$\frac{N (^{14} C)}{N (^{12} C)} = 1.65 \times 10^{- 13}$

As $N (^{12} C) = N_{0} (^{12} C)$ ,

$\begin{array}{l} \frac{N (^{14} C)}{N_{0} (^{14} C)} = \frac{1.65 \times 10^{- 13}}{1.3 \times 10^{- 12}} \\ = 1.27 \times 10^{- 1} \\ = 0.127 \end{array}$

Therefore,

$N (^{14} C) = 0.127 N_{0} (^{14} C)$

Step.2.

Decay equation in terms of half-life is,

$\begin{array}{l} N = N_{0} {(\frac{1}{2})}^{\frac{t}{t_{1}}} \\ {(\frac{1}{2})}^{\frac{t}{t_{1}}} = \frac{N}{N_{0}} \end{array}$

Here, $N$ is the present number of $^{14} C$ atoms, $N_{0}$ represent the initial number of atoms $^{14} C$ atoms and $t_{\frac{1}{2}}$ represents the half-life of $^{14} C$ .

Substitute 5730 years for $t_{\frac{1}{2}}$ and 0.127 for $\frac{N}{N_{0}}$ ,

${(\frac{1}{2})}^{\frac{1}{5730 years}} = 0.127$

$(0.5) \frac{t}{5730 years} = 0.127$

Take natural log on both sides,

$\begin{array}{l} \frac{t}{5730 years} \ln (0.5) = \ln (0.127) \\ t = \frac{\ln (0.127)}{\ln (0.5)} (5730 years) \\ = (2.9771) (5730 years) \\ = 17, 058.783 years \end{array}$

Step.3.

Hence, the age of bone is 17,058.783 years.

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What is the age in years of a bone in which the $^{14} C^{12} C$ ratio is measured to be $1.65 \times 10^{- 13}$ ?

Short Answer

Step by step solution

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What is the age in years of a bone in which the 14C12Cratio is measured to be 1.65×10−13?

Short Answer

Step by step solution

Step.1.

Step.2.

Step.3.

One App. One Place for Learning.

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What is the age in years of a bone in which the $^{14} C^{12} C$ ratio is measured to be $1.65 \times 10^{- 13}$ ?