Question
The half-life of a radioactive kind of strontium is 10 hours. How much will be left after 40 hours, if you start with 80 grams of it?
Answers
Answered by
GPT-4o mini
The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life of strontium is 10 hours.
To find out how much of the substance remains after a certain period, you can use the formula:
\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]
where:
- \(N\) is the amount remaining after time \(t\),
- \(N_0\) is the initial amount,
- \(t\) is the total time elapsed,
- \(t_{1/2}\) is the half-life.
In this problem, we have:
- \(N_0 = 80\) grams,
- \(t = 40\) hours,
- \(t_{1/2} = 10\) hours.
First, find out how many half-lives have passed in 40 hours:
\[
\text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{40 \text{ hours}}{10 \text{ hours}} = 4.
\]
Now, substitute the values into the formula to find \(N\):
\[
N = 80 \left( \frac{1}{2} \right)^{4} = 80 \left( \frac{1}{16} \right) = \frac{80}{16} = 5 \text{ grams}.
\]
Therefore, after 40 hours, there will be **5 grams** of strontium left.
To find out how much of the substance remains after a certain period, you can use the formula:
\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]
where:
- \(N\) is the amount remaining after time \(t\),
- \(N_0\) is the initial amount,
- \(t\) is the total time elapsed,
- \(t_{1/2}\) is the half-life.
In this problem, we have:
- \(N_0 = 80\) grams,
- \(t = 40\) hours,
- \(t_{1/2} = 10\) hours.
First, find out how many half-lives have passed in 40 hours:
\[
\text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{40 \text{ hours}}{10 \text{ hours}} = 4.
\]
Now, substitute the values into the formula to find \(N\):
\[
N = 80 \left( \frac{1}{2} \right)^{4} = 80 \left( \frac{1}{16} \right) = \frac{80}{16} = 5 \text{ grams}.
\]
Therefore, after 40 hours, there will be **5 grams** of strontium left.