Let's analyze the provided data to solve the problem step-by-step.
To determine the half-life of each material, we first use the principle that the mass of a decaying radioactive material after a certain time follows the formula:
[tex]\[ \text{remaining\_mass} = \text{original\_mass} \times (0.5)^{\frac{\text{time}}{\text{half\_life}}} \][/tex]
We are given:
- For Material 1:
- Original mass = 100 grams
- Remaining mass after 21.6 seconds = 12.5 grams
- For Material 2:
- Original mass = 200 grams
- Remaining mass after 21.6 seconds = 25 grams
To isolate the half-life in each case, we rearrange the equation:
[tex]\[ \frac{\text{remaining\_mass}}{\text{original\_mass}} = (0.5)^{\frac{\text{time}}{\text{half\_life}}} \][/tex]
Taking the natural logarithm (or base-2 logarithm) on both sides transforms the equation:
[tex]\[ \log_{2}\left(\frac{\text{remaining\_mass}}{\text{original\_mass}}\right) = \frac{\text{time}}{\text{half\_life}} \cdot \log_{2}(0.5) \][/tex]
Since [tex]\(\log_{2}(0.5) = -1\)[/tex], the equation simplifies to:
[tex]\[ \log_{2}\left(\frac{\text{remaining\_mass}}{\text{original\_mass}}\right) = -\frac{\text{time}}{\text{half\_life}} \][/tex]
[tex]\[ \text{half\_life} = \frac{\text{time}}{-\log_{2}\left(\frac{\text{remaining\_mass}}{\text{original\_mass}}\right)} \][/tex]
[tex]\[ \text{half\_life} = \frac{\text{time}}{\log_{2}\left(\frac{\text{original\_mass}}{\text{remaining\_mass}}\right)} \][/tex]
For Material 1:
[tex]\[ \text{half\_life\_1} = \frac{21.6}{\log_{2}\left(\frac{100}{12.5}\right)} \][/tex]
For Material 2:
[tex]\[ \text{half\_life\_2} = \frac{21.6}{\log_{2}\left(\frac{200}{25}\right)} \][/tex]
Given the calculations, both half-lives turn out to be:
[tex]\[ \text{half\_life\_1} = 7.2 \text{ seconds} \][/tex]
[tex]\[ \text{half\_life\_2} = 7.2 \text{ seconds} \][/tex]
This implies that:
- The half-life of Material 1 is 7.2 seconds.
- The half-life of Material 2 is 7.2 seconds.
Therefore, the correct conclusion based on this analysis is:
The half-life of Material 1 and Material 2 are equal.
