To determine the half-life of a radioactive substance, we can use the formula that describes the decay of a radioactive substance:
[tex]\[ \text{remaining amount} = \text{initial amount} \times \left(\frac{1}{2}\right)^{\left(\frac{\text{time elapsed}}{\text{half-life}}\right)} \][/tex]
Given:
- Initial amount, [tex]\( \text{initial amount} = 1.00 \)[/tex] grams
- Remaining amount after 2.0 hours, [tex]\( \text{remaining amount} = 0.25 \)[/tex] grams
- Time elapsed, [tex]\( \text{time elapsed} = 2.0 \)[/tex] hours
We need to find the half-life of this substance.
1. Start with the provided formula:
[tex]\[ 0.25 = 1.00 \times \left(\frac{1}{2}\right)^{\left(\frac{2.0}{\text{half-life}}\right)} \][/tex]
2. Rewrite the equation in a simpler form:
[tex]\[ 0.25 = \left(\frac{1}{2}\right)^{\left(\frac{2.0}{\text{half-life}}\right)} \][/tex]
3. Take the natural logarithm on both sides to solve for the half-life:
[tex]\[ \ln(0.25) = \ln\left(\left(\frac{1}{2}\right)^{\left(\frac{2.0}{\text{half-life}}\right)}\right) \][/tex]
4. Use the logarithmic property [tex]\( \ln(a^b) = b \ln(a) \)[/tex]:
[tex]\[ \ln(0.25) = \frac{2.0}{\text{half-life}} \cdot \ln\left(\frac{1}{2}\right) \][/tex]
5. We know that [tex]\( \ln\left(\frac{1}{2}\right) \)[/tex] can be simplified:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2) \][/tex]
6. Substitute [tex]\( \ln(0.25) \)[/tex] and [tex]\( -\ln(2) \)[/tex] into the equation:
[tex]\[ \ln(0.25) = \frac{2.0}{\text{half-life}} \cdot (-\ln(2)) \][/tex]
7. Recognize that [tex]\( \ln(0.25) = \ln\left(\left(\frac{1}{2}\right)^2\right) = 2 \ln\left(\frac{1}{2}\right) = 2 (-\ln(2)) \)[/tex]:
[tex]\[ 2 (-\ln(2)) = -2 \ln(2) \][/tex]
8. Substitute back:
[tex]\[ -2 \ln(2) = -\frac{2.0}{\text{half-life}} \ln(2) \][/tex]
9. Cancel out the common terms:
[tex]\[ -2 = -\frac{2.0}{\text{half-life}} \][/tex]
10. Solve for the half-life:
[tex]\[ \text{half-life} = \frac{2.0}{2} = 1.0 \text{ hour} \][/tex]
Therefore, the half-life of this substance is 1.0 hour.
