Question

\({}^{239}{\rm{Pu}}\) is a nuclear waste by-product with a half-life of \({\rm{24,000 y}}\). What fraction of the \({}^{239}{\rm{Pu}}\) present today will be present in \({\rm{1000 y}}\)?

Short answer

The fraction of \({}^{239}{\rm{Pu}}\) present today that will be present in \({\rm{1000 y}}\) is \({\rm{97\% }}\).

Step-by-step solution

Introduction

The formula to calculate the ratio of the radioactive substance:

\(\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - \lambda }} \cdot {\rm{t}}}}\)

Where\({\rm{t}}\)is time,\({\rm{\lambda }}\)is decay constant,\({{\rm{N}}_{\rm{0}}}\)is initial concentration, and\({\rm{N}}\)is the concentration at time\({\rm{t}}\).

From the half-life, calculate\({\rm{\lambda }}\):

\(\begin{array}{l}{\rm{\lambda = }}\frac{{{\rm{ln(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\\{\rm{\lambda = }}\frac{{{\rm{ln(2)}}}}{{{\rm{24000 y}}}}\\{\rm{\lambda = 2}}{\rm{.9}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ }}{{\rm{y}}^{{\rm{ - 1}}}}\end{array}\)

Percentage of \({}_{{\rm{94}}}^{{\rm{239}}}{\rm{Pu}}\) remaining in \({\rm{1000 y}}\)

The time given is \({\rm{1000 y}}\).

The ratio of \({{\rm{N}}_{\rm{0}}}\) can be calculated as follows:

\(\begin{array}{c}\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 2}}{\rm{.9}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;}}{{\rm{y}}^{{\rm{ - 1}}}} \cdot {\rm{1000 y}}}}\\\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = 0}}{\rm{.97}} \cdot {\rm{100}}\\\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = 97\% }}\end{array}\)

Therefore, the value of the ratio is obtained as \({\rm{97\% }}\).