Answer :
To determine the appropriate model for the decay of the radioactive dye, we need to focus on the mathematical properties of exponential decay. Exponential decay can be modeled using the general formula [tex]\( N(t) = N_0 e^{-\lambda t} \)[/tex], where [tex]\( N_0 \)[/tex] is the initial amount, [tex]\( N(t) \)[/tex] is the amount remaining after time [tex]\( t \)[/tex], and [tex]\( \lambda \)[/tex] is the decay constant.
Given:
- Initial amount, [tex]\( N_0 = 13 \)[/tex] milligrams
- Amount after 12 minutes, [tex]\( N = 4.75 \)[/tex] milligrams
- Time, [tex]\( t = 12 \)[/tex] minutes
We use the exponential decay formula to find the decay constant [tex]\( \lambda \)[/tex]:
[tex]\[ N = N_0 e^{-\lambda t} \][/tex]
Plugging in the given values:
[tex]\[ 4.75 = 13 e^{-\lambda \cdot 12} \][/tex]
To isolate [tex]\( \lambda \)[/tex], we begin by dividing both sides by 13:
[tex]\[ \frac{4.75}{13} = e^{-12\lambda} \][/tex]
Taking the natural logarithm ln of both sides:
[tex]\[ \ln \left( \frac{4.75}{13} \right) = -12\lambda \][/tex]
Solving for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = -\frac{\ln \left( \frac{4.75}{13} \right)}{12} \][/tex]
After calculating, we find:
[tex]\[ \lambda \approx 0.0839 \][/tex]
Now, let's evaluate each model option:
- Option (a) [tex]\( f(t) = 13(0.0805)^t \)[/tex]: This model uses a power function [tex]\( (0.0805)^t \)[/tex] which is not suitable for exponential decay as exponential decay should take the form [tex]\( e^{-\lambda t} \)[/tex].
- Option (b) [tex]\( f(t) = 13 e^{0.9195 t} \)[/tex]: This model features a positive exponent [tex]\( 0.9195 \)[/tex]. A positive exponent indicates exponential growth rather than decay.
- Option (c) [tex]\( f(t) = 13 e^{(-0.0839 t)} \)[/tex]: This model is in the correct form for exponential decay. The calculated value [tex]\( \lambda \approx 0.0839 \)[/tex] closely matches the exponent in this model, validating its appropriateness.
- Option (d) [tex]\( f(t) = \frac{4.75}{1 + 13 e^{-0.83925 t}} \)[/tex]: This model represents a logistic function, which is typically used for population growth or decay with a carrying capacity, not for simple exponential decay.
Therefore, the appropriate model for the exponential decay of the radioactive dye is option (c):
[tex]\[ f(t)= 13 e^{(-0.0839 t)} \][/tex]
This model properly represents the decay process with the correct decay constant reflecting the problem's conditions.
Given:
- Initial amount, [tex]\( N_0 = 13 \)[/tex] milligrams
- Amount after 12 minutes, [tex]\( N = 4.75 \)[/tex] milligrams
- Time, [tex]\( t = 12 \)[/tex] minutes
We use the exponential decay formula to find the decay constant [tex]\( \lambda \)[/tex]:
[tex]\[ N = N_0 e^{-\lambda t} \][/tex]
Plugging in the given values:
[tex]\[ 4.75 = 13 e^{-\lambda \cdot 12} \][/tex]
To isolate [tex]\( \lambda \)[/tex], we begin by dividing both sides by 13:
[tex]\[ \frac{4.75}{13} = e^{-12\lambda} \][/tex]
Taking the natural logarithm ln of both sides:
[tex]\[ \ln \left( \frac{4.75}{13} \right) = -12\lambda \][/tex]
Solving for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = -\frac{\ln \left( \frac{4.75}{13} \right)}{12} \][/tex]
After calculating, we find:
[tex]\[ \lambda \approx 0.0839 \][/tex]
Now, let's evaluate each model option:
- Option (a) [tex]\( f(t) = 13(0.0805)^t \)[/tex]: This model uses a power function [tex]\( (0.0805)^t \)[/tex] which is not suitable for exponential decay as exponential decay should take the form [tex]\( e^{-\lambda t} \)[/tex].
- Option (b) [tex]\( f(t) = 13 e^{0.9195 t} \)[/tex]: This model features a positive exponent [tex]\( 0.9195 \)[/tex]. A positive exponent indicates exponential growth rather than decay.
- Option (c) [tex]\( f(t) = 13 e^{(-0.0839 t)} \)[/tex]: This model is in the correct form for exponential decay. The calculated value [tex]\( \lambda \approx 0.0839 \)[/tex] closely matches the exponent in this model, validating its appropriateness.
- Option (d) [tex]\( f(t) = \frac{4.75}{1 + 13 e^{-0.83925 t}} \)[/tex]: This model represents a logistic function, which is typically used for population growth or decay with a carrying capacity, not for simple exponential decay.
Therefore, the appropriate model for the exponential decay of the radioactive dye is option (c):
[tex]\[ f(t)= 13 e^{(-0.0839 t)} \][/tex]
This model properly represents the decay process with the correct decay constant reflecting the problem's conditions.