Answer :
To find the correct equation and approximate the remaining mass of a 200 mg sample of iron after 12 years, follow these steps:
1. Understand the Given Function:
The function to determine the remaining mass of a radioactive substance after [tex]\( t \)[/tex] years is provided:
[tex]\[ f(t) = m \left( 0.5 \right)^{\frac{t}{h}} \][/tex]
Where [tex]\( m \)[/tex] is the initial mass, and [tex]\( h \)[/tex] is the half-life in years.
2. Identify Given Values:
- The initial mass [tex]\( m = 200 \)[/tex] mg.
- The half-life of iron [tex]\( h = 2.7 \)[/tex] years.
- The time duration [tex]\( t = 12 \)[/tex] years.
3. Substitute the Values into the Function:
Substitute [tex]\( m = 200 \)[/tex], [tex]\( h = 2.7 \)[/tex], and [tex]\( t = 12 \)[/tex] into the function:
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{12}{2.7}} \][/tex]
4. Calculate the Exponent:
[tex]\[ \frac{12}{2.7} \approx 4.444 \][/tex]
5. Evaluate the Expression:
Evaluate [tex]\( (0.5)^{4.444} \)[/tex]:
[tex]\[ (0.5)^{4.444} \approx 0.045929 \][/tex]
6. Multiply by the Initial Mass:
Multiply the resulting value by the initial mass:
[tex]\[ 200 \times 0.045929 \approx 9.1858 \][/tex]
7. Compare With Provided Choices:
From the available choices:
- [tex]\( f(t)=2.7(0.5)^{\frac{t}{200}} ; 2.6 \text{ mg} \)[/tex]
- [tex]\( f(t)=2.7(0.5)^t ; 0.0007 \text{ mg} \)[/tex]
- [tex]\( f(t)=200(0.5)^t ; 0.05 \text{ mg} \)[/tex]
- [tex]\( f(t)=200(0.5)^{\frac{t}{2.7}} ; 9.2 \text{ mg} \)[/tex]
The equation
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{t}{2.7}} \][/tex]
correctly incorporates the initial mass, the half-life, and the time. The remaining mass calculation also closely matches the provided approximation of 9.2 mg for the 12-year duration.
Therefore, the correct equation is:
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{t}{2.7}} \][/tex]
And the approximate remaining mass after 12 years is:
[tex]\[ \approx 9.2 \text{ mg} \][/tex]
Thus, the answer is:
[tex]\[ f(t)=200(0.5)^{\frac{t}{2.7}} ; 9.2 \text{ mg} \][/tex]
1. Understand the Given Function:
The function to determine the remaining mass of a radioactive substance after [tex]\( t \)[/tex] years is provided:
[tex]\[ f(t) = m \left( 0.5 \right)^{\frac{t}{h}} \][/tex]
Where [tex]\( m \)[/tex] is the initial mass, and [tex]\( h \)[/tex] is the half-life in years.
2. Identify Given Values:
- The initial mass [tex]\( m = 200 \)[/tex] mg.
- The half-life of iron [tex]\( h = 2.7 \)[/tex] years.
- The time duration [tex]\( t = 12 \)[/tex] years.
3. Substitute the Values into the Function:
Substitute [tex]\( m = 200 \)[/tex], [tex]\( h = 2.7 \)[/tex], and [tex]\( t = 12 \)[/tex] into the function:
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{12}{2.7}} \][/tex]
4. Calculate the Exponent:
[tex]\[ \frac{12}{2.7} \approx 4.444 \][/tex]
5. Evaluate the Expression:
Evaluate [tex]\( (0.5)^{4.444} \)[/tex]:
[tex]\[ (0.5)^{4.444} \approx 0.045929 \][/tex]
6. Multiply by the Initial Mass:
Multiply the resulting value by the initial mass:
[tex]\[ 200 \times 0.045929 \approx 9.1858 \][/tex]
7. Compare With Provided Choices:
From the available choices:
- [tex]\( f(t)=2.7(0.5)^{\frac{t}{200}} ; 2.6 \text{ mg} \)[/tex]
- [tex]\( f(t)=2.7(0.5)^t ; 0.0007 \text{ mg} \)[/tex]
- [tex]\( f(t)=200(0.5)^t ; 0.05 \text{ mg} \)[/tex]
- [tex]\( f(t)=200(0.5)^{\frac{t}{2.7}} ; 9.2 \text{ mg} \)[/tex]
The equation
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{t}{2.7}} \][/tex]
correctly incorporates the initial mass, the half-life, and the time. The remaining mass calculation also closely matches the provided approximation of 9.2 mg for the 12-year duration.
Therefore, the correct equation is:
[tex]\[ f(t) = 200 \left( 0.5 \right)^{\frac{t}{2.7}} \][/tex]
And the approximate remaining mass after 12 years is:
[tex]\[ \approx 9.2 \text{ mg} \][/tex]
Thus, the answer is:
[tex]\[ f(t)=200(0.5)^{\frac{t}{2.7}} ; 9.2 \text{ mg} \][/tex]
