Chapter 39, Problem 61A

A product of nuclear power plants is the isotope cesium- 137, which has a half-life of 30 years. How long will it take for this isotope to decay to one-sixteenth its original amount?

To determine

The time for the decay of amount of Cesium-137 to one sixteenth of original amount.

Answer to Problem 61A

The isotope decays to one sixteenth of original amount in $120\text{years}$ .

Explanation of Solution

Given info:

Half-life of Cesium-137, $t_{1/2}=30\text{years}$

Original amount of Cesium-137, $N_0=1\text{unit}$

Final amount of Cesium-137, $N=\frac{1}{16}\text{unit}$

Formula used:

The expression to calculate the number of years for decay is,

$N=N_0{\left(\frac{1}{2}\right)}^{\frac{T}{t_{1/2}}}$

Calculation:

Substitute the values in the above expression.

$\begin{array}{c}\frac{1}{16}\text{unit}=\left(1\text{unit}\right){\left(\frac{1}{2}\right)}^{\frac{T}{t_{1/2}}}\\ {\left(\frac{1}{2}\right)}^{\frac{T}{t_{1/2}}}={\left(\frac{1}{2}\right)}^4\\ \frac{T}{30\text{years}}=4\\ T=120\text{years}\end{array}$

Conclusion:

Thus, the isotope decays to one sixteenth of original amount in $120\text{years}$ .