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

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

Step-by-step explanation:

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

`\displaystyle (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 = (\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) * 22.5\; \text{hours} = 4.5\; \text{hours}`.