Certainly! Let's solve this step-by-step. You are given that a sample of a radioactive substance initially weighs 40 mg and after one year, it weighs 23.5 mg. The aim is to determine the half-life of the substance.
### Step-by-Step Solution:
1. Understand the Radioactive Decay Model:
The decay of a radioactive substance can be modeled using the formula:
[tex]\[
N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}
\][/tex]
where:
- [tex]\(N(t)\)[/tex] is the quantity remaining after time [tex]\(t\)[/tex].
- [tex]\(N_0\)[/tex] is the initial quantity.
- [tex]\(T\)[/tex] is the half-life of the substance.
- [tex]\(t\)[/tex] is the elapsed time.
2. Insert Given Values:
- The initial weight [tex]\(N_0 = 40\)[/tex] mg.
- The weight after one year [tex]\(N(t) = 23.5\)[/tex] mg.
- The time elapsed [tex]\(t = 1\)[/tex] year.
3. Set Up the Equation:
Substituting the known values into the decay formula:
[tex]\[
23.5 = 40 \left(\frac{1}{2}\right)^{\frac{1}{T}}
\][/tex]
Simplifying the equation:
[tex]\[
\frac{23.5}{40} = \left(\frac{1}{2}\right)^{\frac{1}{T}}
\][/tex]
Calculate the fraction:
[tex]\[
0.5875 = \left(\frac{1}{2}\right)^{\frac{1}{T}}
\][/tex]
4. Solve for [tex]\(T\)[/tex]:
To solve for the half-life [tex]\(T\)[/tex], we need to use logarithms. Taking the natural logarithm (ln) of both sides:
[tex]\[
\ln(0.5875) = \ln\left(\left(\frac{1}{2}\right)^{\frac{1}{T}}\right)
\][/tex]
We can use the logarithm property [tex]\(\ln(a^b) = b \cdot \ln(a)\)[/tex]:
[tex]\[
\ln(0.5875) = \frac{1}{T} \cdot \ln\left(\frac{1}{2}\right)
\][/tex]
5. Isolate [tex]\(T\)[/tex]:
Rearranging to solve for [tex]\(T\)[/tex]:
[tex]\[
T = \frac{\ln\left(\frac{1}{2}\right)}{\ln(0.5875)}
\][/tex]
By using the calculated natural logarithms:
- [tex]\(\ln(0.5) \approx -0.693147\)[/tex]
- [tex]\(\ln(0.5875) \approx -0.530628\)[/tex]
6. Calculate [tex]\(T\)[/tex]:
Substituting these values into the equation:
[tex]\[
T = \frac{-0.693147}{-0.530628} \approx 1.303
\][/tex]
### Result:
So, the half-life of the substance is approximately 1.303 years.
