Question
What is the half-life of a radioactive sample that is 75% decayed after 142.8 days?(1 point)
Responses
35.7 days
35.7 days
71.4 days
71.4 days
47.6 days
47.6 days
107.1 days
Responses
35.7 days
35.7 days
71.4 days
71.4 days
47.6 days
47.6 days
107.1 days
Answers
Answered by
GPT-4o mini
To find the half-life of a radioactive sample that is 75% decayed after 142.8 days, we can use the concept of half-lives and the formula for exponential decay.
When a radioactive sample is 75% decayed, it means 25% of the original radioactive substance remains. This can be expressed in terms of half-lives:
If \( N_0 \) is the initial amount, then after one half-life \( N = \frac{N_0}{2} \), after two half-lives \( N = \frac{N_0}{4} \), and after three half-lives \( N = \frac{N_0}{8} \), and so on.
Since we know that 25% of the original amount is left, we realize that this corresponds to the decay occurring through two half-lives because:
- After 1 half-life: 50% remains
- After 2 half-lives: 25% remains (which is 75% decayed)
So, to find the half-life \( t_{1/2} \):
If 2 half-lives correspond to 142.8 days, we can set up the equation:
\[ 2 \cdot t_{1/2} = 142.8 \]
Dividing both sides by 2 gives:
\[ t_{1/2} = \frac{142.8}{2} = 71.4 \text{ days} \]
Thus, the half-life of the radioactive sample is **71.4 days**.
When a radioactive sample is 75% decayed, it means 25% of the original radioactive substance remains. This can be expressed in terms of half-lives:
If \( N_0 \) is the initial amount, then after one half-life \( N = \frac{N_0}{2} \), after two half-lives \( N = \frac{N_0}{4} \), and after three half-lives \( N = \frac{N_0}{8} \), and so on.
Since we know that 25% of the original amount is left, we realize that this corresponds to the decay occurring through two half-lives because:
- After 1 half-life: 50% remains
- After 2 half-lives: 25% remains (which is 75% decayed)
So, to find the half-life \( t_{1/2} \):
If 2 half-lives correspond to 142.8 days, we can set up the equation:
\[ 2 \cdot t_{1/2} = 142.8 \]
Dividing both sides by 2 gives:
\[ t_{1/2} = \frac{142.8}{2} = 71.4 \text{ days} \]
Thus, the half-life of the radioactive sample is **71.4 days**.