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Tju [1.3M]
3 years ago
5

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 ha

lf-life of this material?
Chemistry
1 answer:
Nana76 [90]3 years ago
4 0

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}.

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