Chapter 25.2, Problem 16LC

Interpretation Introduction

Interpretation: To calculate the half-life of cobalt-60.

Concept Introduction: Half-life is the time needed for half the nuclei in a radioactive sample to decay into products. It is denoted by $t_{1/2}$ .

Answer to Problem 16LC

The half-life of cobalt-60 is $5.25\text{ years}$ .

Explanation of Solution

The integrated rate equation can be given as follows for radioactive disintegration.

  $\lambda =\frac{1}{t}\ln \frac{\left[A_{°}\right]}{\left[A\right]}$   ......(1)

Here, $\lambda$ represents the decay constant, $t$ represents the time, $A$ represents the final amount of radioisotope, and $A_{°}$ represents the initial amount of radioisotope.

Substituting $10.5\text{ years}$ for $t$ , $0.800\text{ g}$ for $A$ , and for $0.200\text{ g}$

  $A_{°}$ in equation (1) in order to calculate the decay constant.

  $\begin{array}{c}\lambda =\frac{1}{t}\ln \frac{\left[A_{°}\right]}{\left[A\right]}\\ =\frac{1}{10.5\text{ years}}\ln \frac{0.8\cancel{\text{g}}}{0.2\cancel{\text{g}}}\\ =\frac{1}{10.5}\ln 4\\ =0.132{\text{ years}}^{-1}\end{array}$

Half-life can be expressed as follows:

  $t_{1/2}=\frac{0.693}{\lambda }$   ......(2)

Here, $\lambda$ represents the decay constant and $t_{1/2}$ represents the half-life.

Substituting $0.132{\text{ years}}^{-1}$ for $\lambda$ in equation (2) in order to calculate the half-life.

  $\begin{array}{c}{t}_{1/2}=\frac{0.693}{\lambda }\\ =\frac{0.693}{0.132{\text{ years}}^{-1}}\\ =5.25\text{ years}\end{array}$

Conclusion

Thus, the half-life of cobalt-60 is $5.25\text{ years}$ .