Certainly! Let's solve the problem step by step, considering the half-life property of the substance.
### Key Concepts:
1. Half-Life: The time required for a quantity to reduce to half its initial value.
2. Exponential Decay Formula:
[tex]\[
\text{Remaining Amount} = \text{Initial Amount} \times \left(\frac{1}{2}\right)^{\frac{\text{time}}{\text{half-life}}}
\][/tex]
### Given Data:
1. Initial weight of bromine sample, [tex]\( \text{Initial Amount} = 20 \)[/tex] mg.
2. Half-life of bromine, [tex]\( \text{half-life} = 57 \)[/tex] hours.
We need to find out how much of the sample will remain after various time intervals.
### Time Interval 1: After 57 hours
- Using the exponential decay formula:
[tex]\[
\text{Remaining Amount} = 20 \times \left(\frac{1}{2}\right)^{\frac{57}{57}} = 20 \times \left(\frac{1}{2}\right)^{1} = 20 \times 0.5 = 10 \text{ mg}
\][/tex]
So, after 57 hours, 10 mg of the sample will remain.
### Time Interval 2: After 28.5 hours
- Using the exponential decay formula:
[tex]\[
\text{Remaining Amount} = 20 \times \left(\frac{1}{2}\right)^{\frac{28.5}{57}} = 20 \times \left(\frac{1}{2}\right)^{0.5}
\][/tex]
We know that:
[tex]\[
\left(\frac{1}{2}\right)^{0.5} = \frac{1}{\sqrt{2}} \approx 0.7071
\][/tex]
So:
[tex]\[
\text{Remaining Amount} = 20 \times 0.7071 \approx 14.14 \text{ mg}
\][/tex]
Hence, after 28.5 hours, approximately 14.142 mg of the sample will remain.
### Time Interval 3: After 1 hour
- Using the exponential decay formula:
[tex]\[
\text{Remaining Amount} = 20 \times \left(\frac{1}{2}\right)^{\frac{1}{57}}
\][/tex]
Approximate calculation gives:
[tex]\[
\left(\frac{1}{2}\right)^{\frac{1}{57}} \approx 0.9878
\][/tex]
So:
[tex]\[
\text{Remaining Amount} = 20 \times 0.9878 \approx 19.76 \text{ mg}
\][/tex]
After 1 hour, approximately 19.758 mg of the sample will remain.
### Summary
- After 57 hours: 10 mg
- After 28.5 hours: Approximately 14.142 mg
- After 1 hour: Approximately 19.758 mg
These calculations demonstrate how the weight of the bromine sample decreases over time given its half-life.
