91Ó°ÊÓ

Time after death = Eighty centuries

Mathematical expression for the half-life is given by:

${\mathbf{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.693}}{\mathbf{λ}}$

…………………. (1)

Where, $\mathbf{λ}$= Disintegration constant.

Mathematical expression for the activity of the radioactive sample is given by:

$\mathbf{R}\mathbf{=}\mathbf{λ\pm ·}$ …………………………. (2)

Also, the radioactive decay equation is given by:

role="math" localid="1658467699516" $\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{−}\mathbf{λ³Ù}}$

Where,

role="math" localid="1658467707363" ${\mathbf{N}}_{\mathbf{0}}$= Initial activity when$\mathbf{t}\mathbf{=}\mathbf{0}$

The number of carbon atoms can be calculated by using the following equation as:

${\mathrm{N}}_0=\frac{\mathrm{W}}{\mathrm{M}}{\mathrm{N}}_{\mathrm{A}}$

Substitute$1.0\mathrm{g}$for W,$12\text{ g/³¾´Ç}\mathrm{l}$for$\mathrm{M}$ and$6.023×{10}^{23}$for${\mathrm{N}}_{\mathrm{A}}$in the above equation, we get:

$\begin{array}{c}{\mathrm{N}}_0=\frac{\left(1\text{ g}\right)}{\left(12\text{ g/³¾´Çl}\right)}(6.02×{10}^{23}\text{/mol})\\ =5.02×{10}^{22}\end{array}$

The number of carbon 14 will be:

$\begin{array}{c}\mathrm{N}=5.02×{10}^{22}×1.3×{10}^{−12}\\ =6.52×{10}^{10}\end{array}$

We can get decay constant as:

$\mathrm{λ}=\frac{\text{In}2}{{\mathrm{T}}_{1/2}}$

Substitute$5730\text{ }\mathrm{yr}$for the half-life period of carbon in the above equation, we get:

$\begin{array}{c}\mathrm{λ}=\frac{\mathrm{In}2}{5730\text{ }\mathrm{yr}}\\ =1.21×{10}^{−4}\text{ }{\mathrm{yr}}^{−1}\end{array}$

In eighty centuries, the number of particles will change according :

${\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{−\mathrm{λ³Ù}}$

Substitute $6.52×{10}^{10}$for ${\mathrm{N}}_0$, $1.20×{10}^{−4}{\mathrm{yr}}^{−1}$for $\mathrm{λ}$, and $8000$yr for the time in the above equation, we get,

$\begin{array}{c}{\mathrm{N}}_1=6.52×{10}^{10}{\mathrm{e}}^{−(1.21×{10}^{−4}\text{ }{\mathrm{yr}}^{−1})\left(8000\text{ }\mathrm{yr}\right)}\\ =2.48×{10}^{10}\end{array}$

It’s related to the decay constant by:

${\mathrm{R}}_1={\mathrm{λ\pm ·}}_1$

Substitute $1.20×{10}^{−4}\text{ }y{r}^{−1}$for $\mathrm{λ}$and $2.48×{10}^{10}$for ${\mathrm{N}}_1$in the above equation, we get

$\begin{array}{c}{\mathrm{R}}_1=(1.21×{10}^{−4}{\text{ y°ù}}^{−1})(2.48×{10}^{10})\\ =(3.00×{10}^6{\text{ y°ù}}^{−\text{1}})\left(\frac{\text{1 y±ð²¹°ù}}{\text{365 d²¹²â²õ}}\right)\left(\frac{\text{1 d²¹²â}}{\text{24 h°ù}}\right)\left(\frac{\text{1 h°ù}}{\text{60″¾¾±²Ô}}\right)\\ =5.7\text{ }{\min }^{−1}\end{array}$

Thus, the decay rate is $5.7\text{ }{\min }^{−1}$.