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To find the half-life of a substance, we need to determine the time it takes for half of the initial mass to decay. Let's follow the detailed, step-by-step process to find the half-life, [tex]\( T_{1/2} \)[/tex], given the information provided.
### Step-by-Step Solution:
1. Understand the Decay Formula:
The given decay formula is:
[tex]\[ N = N_0 e^{-k t} \][/tex]
where
- [tex]\( N_0 \)[/tex] is the initial mass,
- [tex]\( N \)[/tex] is the mass at time [tex]\( t \)[/tex],
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time.
2. Define Half-Life:
The half-life ([tex]\( T_{1/2} \)[/tex]) is the time required for the substance to decay to half of its initial mass, i.e., when [tex]\( N = \frac{N_0}{2} \)[/tex].
3. Set up the Half-Life Equation:
Substitute [tex]\( N = \frac{N_0}{2} \)[/tex] into the decay formula:
[tex]\[ \frac{N_0}{2} = N_0 e^{-k t_{1/2}} \][/tex]
Divide both sides by [tex]\( N_0 \)[/tex] to isolate the exponential term:
[tex]\[ \frac{1}{2} = e^{-k t_{1/2}} \][/tex]
4. Take the Natural Logarithm on Both Sides:
Apply the natural logarithm (ln) to both sides to solve for [tex]\( t_{1/2} \)[/tex]:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln\left(e^{-k t_{1/2}}\right) \][/tex]
Using the property of logarithms [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln\left(\frac{1}{2}\right) = -k t_{1/2} \][/tex]
5. Solve for [tex]\( t_{1/2} \)[/tex]:
Rearrange the equation to solve for [tex]\( t_{1/2} \)[/tex]:
[tex]\[ t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-k} \][/tex]
Using the fact that [tex]\( \ln\left(\frac{1}{2}\right) = -\ln(2) \)[/tex]:
[tex]\[ t_{1/2} = \frac{-\ln(2)}{-k} \][/tex]
Simplifying further:
[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]
6. Substitute the Given Decay Constant:
Given [tex]\( k = 0.1024 \)[/tex], substitute this value into the equation:
[tex]\[ t_{1/2} = \frac{\ln(2)}{0.1024} \][/tex]
We know that [tex]\( \ln(2) \approx 0.693 \)[/tex]:
[tex]\[ t_{1/2} = \frac{0.693}{0.1024} \approx 6.7690154351557155 \][/tex]
7. Round to the Nearest Tenth:
Finally, round the result to the nearest tenth:
[tex]\[ t_{1/2} \approx 6.8 \][/tex]
### Conclusion:
The half-life of the substance is approximately [tex]\( 6.8 \)[/tex] days.
### Step-by-Step Solution:
1. Understand the Decay Formula:
The given decay formula is:
[tex]\[ N = N_0 e^{-k t} \][/tex]
where
- [tex]\( N_0 \)[/tex] is the initial mass,
- [tex]\( N \)[/tex] is the mass at time [tex]\( t \)[/tex],
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time.
2. Define Half-Life:
The half-life ([tex]\( T_{1/2} \)[/tex]) is the time required for the substance to decay to half of its initial mass, i.e., when [tex]\( N = \frac{N_0}{2} \)[/tex].
3. Set up the Half-Life Equation:
Substitute [tex]\( N = \frac{N_0}{2} \)[/tex] into the decay formula:
[tex]\[ \frac{N_0}{2} = N_0 e^{-k t_{1/2}} \][/tex]
Divide both sides by [tex]\( N_0 \)[/tex] to isolate the exponential term:
[tex]\[ \frac{1}{2} = e^{-k t_{1/2}} \][/tex]
4. Take the Natural Logarithm on Both Sides:
Apply the natural logarithm (ln) to both sides to solve for [tex]\( t_{1/2} \)[/tex]:
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln\left(e^{-k t_{1/2}}\right) \][/tex]
Using the property of logarithms [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln\left(\frac{1}{2}\right) = -k t_{1/2} \][/tex]
5. Solve for [tex]\( t_{1/2} \)[/tex]:
Rearrange the equation to solve for [tex]\( t_{1/2} \)[/tex]:
[tex]\[ t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-k} \][/tex]
Using the fact that [tex]\( \ln\left(\frac{1}{2}\right) = -\ln(2) \)[/tex]:
[tex]\[ t_{1/2} = \frac{-\ln(2)}{-k} \][/tex]
Simplifying further:
[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]
6. Substitute the Given Decay Constant:
Given [tex]\( k = 0.1024 \)[/tex], substitute this value into the equation:
[tex]\[ t_{1/2} = \frac{\ln(2)}{0.1024} \][/tex]
We know that [tex]\( \ln(2) \approx 0.693 \)[/tex]:
[tex]\[ t_{1/2} = \frac{0.693}{0.1024} \approx 6.7690154351557155 \][/tex]
7. Round to the Nearest Tenth:
Finally, round the result to the nearest tenth:
[tex]\[ t_{1/2} \approx 6.8 \][/tex]
### Conclusion:
The half-life of the substance is approximately [tex]\( 6.8 \)[/tex] days.
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