91Ó°ÊÓ

It’s given that t = 200,000 y

As we know, the Mathematical expression for the half-life is given by:

$t_{\frac{1}{2}}=\frac{0.693}{\mathrm{λ}}$ ………….. (1)

Where $\mathrm{λ}$disintegration constant.

Expression for the activity of radioactivity sample is given by:

$R=\mathrm{λ}N$ ………………. (2)

And radioactive decay equation is given by:

$N=N_0{e}^{-\mathrm{λ}t}$

The number of carbon atoms can be calculated by using the following question

$N_0=\frac{w}{M}{N}_A$

Out of$\frac{\text{1}}{\text{10}}$mol of carbon atoms can be calculated by using the following equation.

Substitute 1.0 g for w, 10g/mol for M and $\text{6.023}×{\text{10}}^{\text{23}}$for${\text{N}}_{\text{A}}$in the above equation, and we get,

$\begin{array}{rcl}{N}_0& =& \frac{1g}{10g/mol}(6.02×{10}^{23}/mol)\\ & =& 5.02×{10}^{22}\\ & & \end{array}$

The number of atoms is:

$\begin{array}{rcl}N& =& 6.02×{10}^{22}×1.3×{10}^{-12}\\ & =& 7.83×{10}^{10}\\ & & \end{array}$

The relation between half-life and decay constant as:

$\mathrm{λ}=\frac{In2}{T_{1/2}}$

Substitute 5730 y for${\text{T}}_{\text{1/2}}$ in the above equation, we get:

$\begin{array}{rcl}\mathrm{λ}& =& \frac{In2}{5730y}\\ & =& 1.21×{10}^{-4}{y}^{-1}\\ & & \end{array}$

In 200,000 y, the number of the particle will change according to:

$\begin{array}{rcl}{N}_1& =& N_0{e}^{-\mathrm{λ}t}\\ & =& 7.83×{10}^{10}{e}^{-1.21×{10}^{-4}{y}^{-1}×200,000y}\\ & =& 2.42\\ & & \end{array}$

There are $\text{7.83}×{\text{10}}^{\text{10}}$particles of carbon-14 datingto predict the age of old bone, so around 2.42 atoms will be left.

Carbon-14 is not particularly good for dating anything this old since the carbon-14 atoms left are a relatively small fraction of the original atoms, making measurements difficult.