Question

Eighty centuries after its death, what will be the decay rate of 1g of carbon from the thigh bone of an animal?

Short answer

The decay rate will be $5.7{\min }^{-1}$.

Step-by-step solution

Given information

Time after death = Eighty centuries

Mathematical expression for the half-life is given by:

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

…………………. (1)

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

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

$\mathbf{R}\mathbf{=}\mathbf{\lambda N}$ …………………………. (2)

Also, the radioactive decay equation is given by:

role="math" localid="1658467699516" $\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda t}}$

Where,

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

Calculation for 1g of carbon

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,$12\text{\hspace{0.17em}g/mo}\mathrm{l}$for$\mathrm{M}$ and$6.023\times {10}^{23}$for${\mathrm{N}}_{\mathrm{A}}$in the above equation, we get:

$\begin{array}{c}{\mathrm{N}}_0=\frac{\left(1\text{\hspace{0.17em}g}\right)}{\left(12\text{\hspace{0.17em}g/mol}\right)}(6.02\times {10}^{23}\text{/mol})\\ =5.02\times {10}^{22}\end{array}$

The number of carbon 14 will be:

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

We can get decay constant as:

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

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

$\begin{array}{c}\mathrm{\lambda }=\frac{\mathrm{In}2}{5730\mathrm{yr}}\\ =1.21\times {10}^{-4}{\mathrm{yr}}^{-1}\end{array}$

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

${\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$

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

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

Decay rate

It’s related to the decay constant by:

${\mathrm{R}}_1={\mathrm{\lambda N}}_1$

Substitute $1.20\times {10}^{-4}y{r}^{-1}$for $\lambda$and $2.48\times {10}^{10}$for ${\mathrm{N}}_1$in the above equation, we get

$\begin{array}{c}{\mathrm{R}}_1=(1.21\times {10}^{-4}{\text{\hspace{0.17em}yr}}^{-1})(2.48\times {10}^{10})\\ =(3.00\times {10}^6{\text{\hspace{0.17em}yr}}^{-\text{1}})\left(\frac{\text{1\hspace{0.17em}year}}{\text{365\hspace{0.17em}days}}\right)\left(\frac{\text{1\hspace{0.17em}day}}{\text{24\hspace{0.17em}hr}}\right)\left(\frac{\text{1\hspace{0.17em}hr}}{\text{60\hspace{0.17em}min}}\right)\\ =5.7{\min }^{-1}\end{array}$

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