Question

The half-life of a radioactive substance is 5730 years. If carbon-14 decays and only 80% of it remains after a certain period, what is the time elapsed (in years)?

• A. \( t = \frac{5730 \times 100}{0.3} \log \left(\frac{100}{80}\right) \)
• B. \( t = \frac{5730}{0.3} \log \left(\frac{100}{80}\right) \)
• C. \( t = \frac{5730}{0.3} \log \left(\frac{100}{80}\right) \)
• D. \( t = \frac{5730}{0.3} \log 20 \)

Hint:

In radioactive decay, the half-life formula helps calculate the time elapsed based on the remaining substance and the half-life value.

Answer 1

The Correct Option is B

Step 1: Understanding the Formula.
For a radioactive substance, the formula for half-life is given by: \[ t = \frac{t_{1/2}}{n} \log \left(\frac{100}{\text{remaining percentage}}\right) \] Where \( t_{1/2} = 5730 \, \text{years} \) and the remaining percentage is 80%.
Step 2: Calculation.
Substitute the values into the formula: \[ t = \frac{5730}{0.3} \log \left(\frac{100}{80}\right) = 5730 \times \frac{1}{0.3} \times \log \left( \frac{100}{80} \right) \] Thus, the correct answer is (B).