To determine how long it will take for [tex]\( \frac{1}{16} \)[/tex] of a given amount of radium-226 to remain, considering its half-life is 1599 years, we will use the principles of radioactive decay. Let's go through the solution step-by-step:
1. Understanding Half-Life Concept:
- The half-life ([tex]\( T \)[/tex]) of a radioactive substance is the time it takes for half of the initial amount to decay.
- For radium-226, the half-life is given as 1599 years.
2. Setting Up the Problem:
- Let's denote the initial amount of radium-226 as [tex]\( N_0 \)[/tex].
- The remaining amount after time [tex]\( t \)[/tex] will be [tex]\( N(t) \)[/tex].
- We know that [tex]\( N(t) \)[/tex] should be [tex]\( \frac{1}{16} \)[/tex] of the initial amount [tex]\( N_0 \)[/tex].
3. Formulating the Decay Equation:
- Radioactive decay follows the formula: [tex]\( N(t) = N_0 \times (1/2)^{(t/T)} \)[/tex].
- Here, [tex]\( N(t) \)[/tex] is the remaining amount, [tex]\( N_0 \)[/tex] is the initial amount, [tex]\( t \)[/tex] is the time elapsed, and [tex]\( T \)[/tex] is the half-life.
4. Plugging in the Known Values:
- We are given:
- [tex]\( N(t) = \frac{1}{16} N_0 \)[/tex]
- [tex]\( T = 1599 \)[/tex] years
- Substitute these into the decay formula:
[tex]\[ \frac{1}{16} N_0 = N_0 \times (1/2)^{(t/1599)} \][/tex]
5. Solving for [tex]\( t \)[/tex]:
- First, we can divide both sides of the equation by [tex]\( N_0 \)[/tex] to simplify:
[tex]\[ \frac{1}{16} = (1/2)^{(t/1599)} \][/tex]
- Rewrite [tex]\( \frac{1}{16} \)[/tex] as a power of [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \frac{1}{16} = (1/2)^4 \][/tex]
- So, we have:
[tex]\[ (1/2)^4 = (1/2)^{(t/1599)} \][/tex]
- Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 4 = \frac{t}{1599} \][/tex]
6. Isolating [tex]\( t \)[/tex]:
- Multiply both sides by 1599 to solve for [tex]\( t \)[/tex]:
[tex]\[ t = 4 \times 1599 \][/tex]
7. Calculating the Final Result:
- Perform the multiplication:
[tex]\[ t = 6396 \][/tex]
Thus, it will take 6396 years for [tex]\( \frac{1}{16} \)[/tex] of the given amount of radium-226 to remain.
