Question

The decay constant of a radioactive substance is 5.2 x 10\(^3\) per year. What is its half-life?

Hint:

Notice the units. The decay constant was given in "per year" (or year\(^{-1}\)), so the half-life will be calculated in "years". Always ensure your units are consistent. A very large decay constant implies a very rapid decay, and thus a very short half-life.

Answer 1

Step 1: Understanding the Concept:
The half-life (\(T_{1/2}\)) of a radioactive substance is the time required for half of the initial number of radioactive nuclei to decay. It is inversely related to the decay constant (\(\lambda\)), which represents the probability of decay per nucleus per unit time.
Step 2: Key Formula or Approach:
The relationship between half-life (\(T_{1/2}\)) and the decay constant (\(\lambda\)) is given by:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \] Using the approximation \(\ln(2) \approx 0.693\), the formula becomes:
\[ T_{1/2} \approx \frac{0.693}{\lambda} \] Step 3: Detailed Explanation:
We are given the decay constant \(\lambda = 5.2 \times 10^3\) per year.
Substitute this value into the half-life formula:
\[ T_{1/2} = \frac{0.693}{5.2 \times 10^3 \text{ year}^{-1}} \] Now, we perform the calculation:
\[ T_{1/2} = \frac{0.693}{5.2} \times 10^{-3} \text{ years} \] \[ T_{1/2} \approx 0.1332 \times 10^{-3} \text{ years} \] Expressing this in standard scientific notation:
\[ T_{1/2} \approx 1.33 \times 10^{-1} \times 10^{-3} \text{ years} \] \[ T_{1/2} \approx 1.33 \times 10^{-4} \text{ years} \] Step 4: Final Answer:
The half-life of the radioactive substance is approximately \(1.33 \times 10^{-4}\) years.