The amount of a radioactive substance remaining after [tex]$t$[/tex] years is given by the function [tex]f(t) = m(0.5)^{\frac{t}{h}}[/tex], where [tex]$m$[/tex] is the initial mass and [tex]$h$[/tex] is the half-life in years. Iron has a half-life of 2.7 years.

Which equation gives the mass of a 200 mg iron sample remaining after [tex]$t$[/tex] years, and approximately how many milligrams remain after 12 years?

A. [tex]f(t) = 2.7(0.5)^{\frac{t}{200}} ; 2.6 \, mg[/tex]
B. [tex]f(t) = 2.7(0.5)^t ; 0.0007 \, mg[/tex]
C. [tex]f(t) = 200(0.5)^t ; 0.05 \, mg[/tex]
D. [tex]f(t) = 200(0.5)^{\frac{t}{2.7}} ; 9.2 \, mg[/tex]



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]