Answer :
Let's solve the problem step-by-step:
1. Understanding the Problem:
- The function \( f(h) = m \left( \frac{1}{2} \right)^k \) describes the remaining mass \( m \) of a radioactive substance after \( h \) half-lives.
- The half-life of iron is 2.7 years.
- We need to find the mass of a \( 200 \, \text{mg} \) sample of iron after \( y \) years, where \( y = 12 \text{ years} \).
2. Calculating the Number of Half-Lives:
- The half-life of iron is 2.7 years.
- To find the number of half-lives (\( k \)) that have passed in 12 years, we use:
[tex]\[ k = \frac{y}{\text{half-life}} = \frac{12}{2.7} \][/tex]
3. Calculating the Mass Remaining:
- The initial mass of the sample (\( m_0 \)) is 200 mg.
- Using the equation for the remaining mass:
[tex]\[ f(y) = m_0 \left( \frac{1}{2} \right)^k \][/tex]
- Plugging in the values:
[tex]\[ f(12) = 200 \left( \frac{1}{2} \right)^{\frac{12}{2.7}} \][/tex]
- Simplify the number of half-lives:
[tex]\[ k = \frac{12}{2.7} \approx 4.4444 \][/tex]
- Calculating the remaining mass:
[tex]\[ f(12) \approx 200 \left( \frac{1}{2} \right)^{4.4444} \][/tex]
[tex]\[ f(12) \approx 200 \times 0.04593 \approx 9.2 \, \text{mg} \][/tex]
4. Selecting the Correct Option:
- Given options:
- \( f(x) = 2.7(0.5)^{200}; 1.7 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.5)^{12}; 30.8 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.185)^{12}; 3.2 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.774)^{12}; 9.2 \, \text{mg} \): Correct mass but incorrect base.
None of the given equations match the exact form needed. The correct calculation based on the correct parameters (half-life 2.7 years, initial mass 200 mg) results in approximately 9.2 mg of iron remaining after 12 years.
Given this analysis, the closest answer by value is:
[tex]\[ f(x) = 200(0.774)^{12}; 9.2 \, \text{mg} \][/tex]
However, the correct mathematical representation involves the half-life formula I demonstrated above.
1. Understanding the Problem:
- The function \( f(h) = m \left( \frac{1}{2} \right)^k \) describes the remaining mass \( m \) of a radioactive substance after \( h \) half-lives.
- The half-life of iron is 2.7 years.
- We need to find the mass of a \( 200 \, \text{mg} \) sample of iron after \( y \) years, where \( y = 12 \text{ years} \).
2. Calculating the Number of Half-Lives:
- The half-life of iron is 2.7 years.
- To find the number of half-lives (\( k \)) that have passed in 12 years, we use:
[tex]\[ k = \frac{y}{\text{half-life}} = \frac{12}{2.7} \][/tex]
3. Calculating the Mass Remaining:
- The initial mass of the sample (\( m_0 \)) is 200 mg.
- Using the equation for the remaining mass:
[tex]\[ f(y) = m_0 \left( \frac{1}{2} \right)^k \][/tex]
- Plugging in the values:
[tex]\[ f(12) = 200 \left( \frac{1}{2} \right)^{\frac{12}{2.7}} \][/tex]
- Simplify the number of half-lives:
[tex]\[ k = \frac{12}{2.7} \approx 4.4444 \][/tex]
- Calculating the remaining mass:
[tex]\[ f(12) \approx 200 \left( \frac{1}{2} \right)^{4.4444} \][/tex]
[tex]\[ f(12) \approx 200 \times 0.04593 \approx 9.2 \, \text{mg} \][/tex]
4. Selecting the Correct Option:
- Given options:
- \( f(x) = 2.7(0.5)^{200}; 1.7 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.5)^{12}; 30.8 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.185)^{12}; 3.2 \, \text{mg} \): Incorrect equation and answer.
- \( f(x) = 200(0.774)^{12}; 9.2 \, \text{mg} \): Correct mass but incorrect base.
None of the given equations match the exact form needed. The correct calculation based on the correct parameters (half-life 2.7 years, initial mass 200 mg) results in approximately 9.2 mg of iron remaining after 12 years.
Given this analysis, the closest answer by value is:
[tex]\[ f(x) = 200(0.774)^{12}; 9.2 \, \text{mg} \][/tex]
However, the correct mathematical representation involves the half-life formula I demonstrated above.
