Chapter 25.2, Problem 9SP

Interpretation Introduction

Interpretation: To calculate the mass of manganese-56 at the end.

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 9SP

The remaining mass of manganese-56 is $0.063\text{ mg}$ .

Explanation of Solution

The half-life is 2.6 h.

The time is 10.4 h.

Thus, the number of half-lives can be calculated as follows:

  $\begin{array}{c}n=\frac{t}{t_{1/2}}\\ =\frac{10.4\cancel{\text{h}}}{2.6\cancel{\text{h}}}\\ =4\end{array}$

Hence, the total number of half-lives is 4.

The equation that is used to calculate the mass of radioisotope that remains in a sample can be given as follows:

  $A=A_{°}\times {\left(\frac{1}{2}\right)}^n$   ...... (1)

Here, $A$ represents the remaining amount of radioisotope, $A_{°}$ represents the initial amount of radioisotope, and

  $n$ represents the number of half-lives.

Substituting $1.0\text{ mg}$ for $A_{°}$ and $4$ for $n$ in equation (1) in order to calculate the remaining mass of the radioisotope.

  $\begin{array}{c}A=A_{°}\times {\left(\frac{1}{2}\right)}^n\\ =1.0\text{ mg }{\left(\frac{1}{2}\right)}^4\\ =0.063\text{ mg}\end{array}$

Conclusion

Thus, the remaining mass of manganese-56 is $0.063\text{ mg}$ .