Question

Given initially 100 g of Plutonium-239, how muchtime must pass for amount to drop to$\mathbf{1}\mathbf{\text{\hspace{0.17em}g}}$ ?

Short answer

The time taken is $160.18\text{\hspace{0.17em}thousand\hspace{0.17em}years}$.

Step-by-step solution

Given data

N₀= $100\text{\hspace{0.17em}g}$

N= $1\text{\hspace{0.17em}g}$

Concept of Number of Nuclei which remain

The number of nuclei which remain after time $\mathit{t}$ is given by:

$\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda t}}$ .

Where $\mathbf{\lambda }\mathbf{=}\mathbf{\text{Decay\hspace{0.17em}constant}}\mathbf{=}\frac{\mathbf{ln}\mathbf{2}}{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}$ , ${\mathit{N}}_{\mathbf{0}}$ is the original number of nuclei, ${\mathit{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}$ half-life and $\mathit{N}\mathbf{=}$is the number of nuclei at time t.

Calculate the time required from the Concept of the Number of Nuclei which remain

Plutonium 239's half-life is $24.11$thousand years.

For nuclear decay, obtain:

$\begin{array}{c}\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}\\ ={\mathrm{N}}_0{\mathrm{e}}^{-\left(\frac{\ln 2}{{\mathrm{T}}_{1/2}}\right)\mathrm{t}}\end{array}$

Apply logarithm on both sides:

$\ln \frac{\mathrm{N}}{{\mathrm{N}}_0}=-\frac{\ln 2\times \mathrm{t}}{{\mathrm{T}}_{1/2}}$

Rearrange the above equation for $t$, substitute the values as:

$\begin{array}{c}\mathrm{t}=-\frac{{\mathrm{T}}_{1/2}}{\ln 2}\ln \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)\\ =-\left(\frac{24.11\text{\hspace{0.17em}thousand\hspace{0.17em}years}}{\ln 2}\right)\ln \left(\frac{1\text{\hspace{0.17em}g}}{100\text{\hspace{0.17em}g}}\right)\\ =160.18\text{\hspace{0.17em}thousand\hspace{0.17em}years}\end{array}$

Thus, the time taken is $160.18\text{\hspace{0.17em}thousand\hspace{0.17em}years}$.