Question
The half-life of a radioactive nucleus is 17 seconds. What is the decay constant of this source?
~ 0.020
~ 0.041
~0.017
~ 0.082
~ 0.020
~ 0.041
~0.017
~ 0.082
Answers
Answered by
GPT-4o mini
The decay constant (\( \lambda \)) can be calculated using the relationship between the half-life (\( t_{1/2} \)) and the decay constant given by the formula:
\[
\lambda = \frac{\ln(2)}{t_{1/2}}
\]
Where:
- \( \ln(2) \) is approximately 0.693.
- \( t_{1/2} \) is the half-life.
Given that the half-life is 17 seconds, we can substitute this value into the formula:
\[
\lambda = \frac{0.693}{17 \, \text{s}}
\]
\[
\lambda \approx 0.0407 \, \text{s}^{-1}
\]
This value is approximately \( 0.041 \, \text{s}^{-1} \).
Thus, the decay constant of the source is approximately \( 0.041 \, \text{s}^{-1} \).
The correct answer is:
~ 0.041
\[
\lambda = \frac{\ln(2)}{t_{1/2}}
\]
Where:
- \( \ln(2) \) is approximately 0.693.
- \( t_{1/2} \) is the half-life.
Given that the half-life is 17 seconds, we can substitute this value into the formula:
\[
\lambda = \frac{0.693}{17 \, \text{s}}
\]
\[
\lambda \approx 0.0407 \, \text{s}^{-1}
\]
This value is approximately \( 0.041 \, \text{s}^{-1} \).
Thus, the decay constant of the source is approximately \( 0.041 \, \text{s}^{-1} \).
The correct answer is:
~ 0.041