Question

a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the half-life of this material?

Answer 1

Answer:

Approximately $4.5\; \text{hours}$.

Explanation:

Calculate the ratio between the mass of this sample after $22.5\; \text{hours}$ and the initial mass:

$\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125$.

Let $n$ denote the number of half-lives in that $22.5\; \text{hours}$ (where $n\!$ might not necessarily be an integer.) The mass of the sample is supposed to become $(1/2)$ the previous quantity after each half-life. Therefore, if the initial mass of the sample is $1\; \rm g$ (for example,) the mass of the sample after $\! n$ half-lives would be ${(1/2)}^{n}\; \rm g$. Regardless of the initial mass, the ratio between the mass of the sample after $n\!\!$ half-lives and the initial mass should be ${(1/2)}^{n}$.

For this question:

${(1/2)}^{n}} = 0.03125$.

Take the natural logarithm of both sides of this equation to solve for $n$:

$\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)$.

$n\, [\ln(1/2)] = \ln (0.03125)$.

$\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5$.

In other words, there are $5$ half-lives of this sample in $22.5\; \text{hours}$. If the length of each half-life is constant, that length should be $(1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}$.