A radioactive isotope has a decay constant of 10$^{-6}$s$^{-1}$. Calculate its half-life.

Analysis

To solve this problem, we need to find the relationship between the decay constant ($\lambda$) and the half-life ($T_{1/2}$) of a radioactive substance.

Concept

The half-life of a radioactive isotope is the time required for half of the radioactive nuclei in a sample to decay. The mathematical relationship between the half-life ($T_{1/2}$) and the decay constant ($\lambda$) is given by the formula:

Where:

• $\ln(2)$ is the natural logarithm of 2, which is approximately 0.693.
• $\lambda$ is the decay constant.

Solution

Step 1: Identify the given values.

• Decay constant ($\lambda$) = $10^{-6} \text{ s}^{-1}$

Step 2: Apply the formula.

Step 3: Simplify the expression. When you divide by a negative power of 10, it is the same as multiplying by the positive power of 10:

Step 4: Convert to standard form. To express $0.693 \times 10^6$ in standard scientific notation: Move the decimal point one place to the right:

Conclusion

Comparing our result to the given options, we can see that our calculated value matches option B.

Exam Tip: Always remember that $\ln(2) \approx 0.693$. If you ever forget the formula, just remember that the half-life and decay constant are inversely proportional; a very small decay constant means a very long half-life!