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Juliet started working with a radioactive substance. In 1997, she started with [tex]$200 \, \text{g}$[/tex]. When she measured again in 2002, she had [tex][tex]$100 \, \text{g}$[/tex][/tex] left. In 2007, she had [tex]$50 \, \text{g}$[/tex].

\begin{tabular}{|l|l|}
\hline Radioactive Isotope & \multicolumn{1}{|c|}{Half-life} \\
\hline Rubidium-91 & 58.4 seconds \\
\hline Iodine-131 & 8 days \\
\hline Cobalt-60 & 5 years \\
\hline Carbon-14 & 5730 years \\
\hline Cesium-135 & [tex]$2.3 \times 10^6$[/tex] years \\
\hline
\end{tabular}

Which substance is she most likely measuring?

A. Rubidium-91
B. Iodine-131
C. Cobalt-60
D. Carbon-14

Sagot :

GinnyAnswer
To determine which radioactive substance Juliet is measuring, we need to calculate the half-life of the substance based on the information provided.

Juliet started with [tex]\(200 \, \text{g}\)[/tex] of the substance in 1997. We can use the measured amounts in subsequent years to calculate the half-life.

In 2002 (5 years later), she had [tex]\(100 \, \text{g}\)[/tex] of the substance left. This suggests that the substance halved in 5 years.
[tex]\[ \text{Initial amount} = 200 \, \text{g} \][/tex]
[tex]\[ \text{Amount left in 5 years} = 100 \, \text{g} \][/tex]

Using the half-life formula:
[tex]\[ \text{Amount remaining} = \text{Initial amount} \times \left(\frac{1}{2}\right)^{\frac{\text{time elapsed}}{\text{half-life}}} \][/tex]
Given:
[tex]\[ 100 = 200 \times \left(\frac{1}{2}\right)^{\frac{5 \text{ years}}{\text{half-life}}} \][/tex]

To solve for the half-life:
[tex]\[ \frac{100}{200} = \left(\frac{1}{2}\right)^{\frac{5 \text{ years}}{\text{half-life}}} \][/tex]
[tex]\[ \frac{1}{2} = \left(\frac{1}{2}\right)^{\frac{5 \text{ years}}{\text{half-life}}} \][/tex]

This equation confirms that:
[tex]\[ \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{\frac{5 \text{ years}}{\text{half-life}}} \][/tex]
This simply means:
[tex]\[ \frac{5 \text{ years}}{\text{half-life}} = 1 \][/tex]
Thus:
[tex]\[ \text{half-life} = 5 \text{ years} \][/tex]

We can verify this calculation with the measurement from 2007 (10 years later from 1997), where [tex]\(50 \, \text{g}\)[/tex] of the substance remained:
[tex]\[ \text{Initial amount} = 200 \, \text{g} \][/tex]
[tex]\[ \text{Amount left in 10 years} = 50 \, \text{g} \][/tex]

Using the half-life formula again:
[tex]\[ 50 = 200 \times \left(\frac{1}{2}\right)^{\frac{10 \text{ years}}{\text{half-life}}} \][/tex]

Given:
[tex]\[ \frac{50}{200} = \left(\frac{1}{2}\right)^{\frac{10 \text{ years}}{\text{half-life}}} \][/tex]
[tex]\[ \frac{1}{4} = \left(\frac{1}{2}\right)^{2} \][/tex]

Since:
[tex]\[ \left(\frac{1}{2}\right)^{2} = \frac{1}{2 \times 2} = \frac{1}{4} \][/tex]
This confirms that:
[tex]\[ \text{half-life} = 5 \text{ years} \][/tex]

Comparing this calculated half-life with the given half-lives of various isotopes:
- Rubidium-91: 58.4 seconds
- Iodine-131: 8 days
- Cobalt-60: 5 years
- Carbon-14: 5730 years
- Cesium-135: [tex]\(2.3 \times 10^6\)[/tex] years

The half-life that matches our calculated half-life of 5 years is that of Cobalt-60.

Therefore, the substance Juliet is most likely measuring is Cobalt-60.
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