Question
Lead-202 has a half-life of 53,000 years. How long will it take for 15/16 of a sample of lead-202 to decay?
106,000 years
159,000 years
212,000 years
265,000 years
106,000 years
159,000 years
212,000 years
265,000 years
Answers
GPT-4o mini
To determine how long it takes for \( \frac{15}{16} \) of a sample of Lead-202 to decay, we first need to understand what this means in terms of the remaining fraction of the sample.
If \( \frac{15}{16} \) of the sample has decayed, then \( \frac{1}{16} \) remains.
The decay of a radioactive substance can be described using the half-life formula. The remaining quantity of a radioactive substance after a certain amount of time can be described by the formula:
\[
N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]
Where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life.
In this case, we want \( N(t) = \frac{1}{16} N_0 \) and \( T_{1/2} = 53,000 \) years.
Setting up the equation:
\[
\frac{1}{16} N_0 = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Dividing both sides by \( N_0 \):
\[
\frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Now, we can express \( \frac{1}{16} \) in terms of powers of \( \frac{1}{2} \):
\[
\frac{1}{16} = \left( \frac{1}{2} \right)^4
\]
This gives us:
\[
\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Since the bases are the same, we can set the exponents equal to each other:
\[
4 = \frac{t}{53,000}
\]
Now we can solve for \( t \):
\[
t = 4 \times 53,000 = 212,000 \text{ years}
\]
Therefore, the time it takes for \( \frac{15}{16} \) of a sample of Lead-202 to decay is **212,000 years**.
If \( \frac{15}{16} \) of the sample has decayed, then \( \frac{1}{16} \) remains.
The decay of a radioactive substance can be described using the half-life formula. The remaining quantity of a radioactive substance after a certain amount of time can be described by the formula:
\[
N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]
Where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life.
In this case, we want \( N(t) = \frac{1}{16} N_0 \) and \( T_{1/2} = 53,000 \) years.
Setting up the equation:
\[
\frac{1}{16} N_0 = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Dividing both sides by \( N_0 \):
\[
\frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Now, we can express \( \frac{1}{16} \) in terms of powers of \( \frac{1}{2} \):
\[
\frac{1}{16} = \left( \frac{1}{2} \right)^4
\]
This gives us:
\[
\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{\frac{t}{53,000}}
\]
Since the bases are the same, we can set the exponents equal to each other:
\[
4 = \frac{t}{53,000}
\]
Now we can solve for \( t \):
\[
t = 4 \times 53,000 = 212,000 \text{ years}
\]
Therefore, the time it takes for \( \frac{15}{16} \) of a sample of Lead-202 to decay is **212,000 years**.