Chapter 8.8, Problem 1CYP

To determine

To find: the value of constant $k$ if the half-life of plutonium is $24000$ years.

Answer to Problem 1CYP

  $k=-2.89\times {10}^{-5}$ .

Explanation of Solution

Given:

The plutonium- $239$ had half- life of $24,000\text{ year}$ .

Concept used:

The radio activity decay constant $k$ is defined as the natural logarithm of $0.5$ divided by the half- life.

  $k=\frac{\ln \left(.5\right)}{\text{half- life}}.$

Calculation:

The radio activity decay constant $k$ is defined as the natural logarithm of $0.5$ divided by the half- life.

  $k=\frac{\ln \left(.5\right)}{\text{half- life}}.$

According to the given plutonium- $239$ had half- life of $24,000\text{ year}$ .

  $\begin{array}{l}k=\ln \left(.5\right)÷24,000.\\ \Rightarrow k=-0.693147÷24,000.\\ \Rightarrow k=-0.0000289=-2.89\times {10}^{-5}.\\ \Rightarrow k=-2.89\times {10}^{-5}.\end{array}$

Hence, the value of $k=-2.89\times {10}^{-5}$ .