Question

The half-life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be

• A. 20 minutes
• B. 30 minutes
• C. 40 minutes
• D. 25 minutes
• E. 10 minutes

Hint:

You can visualize this: 100% $\rightarrow$ 50% (1st $T_{1/2}$) $\rightarrow$ 25% (2nd $T_{1/2}$) $\rightarrow$ 12.5% (3rd $T_{1/2}$). The jump from 50% remaining to 12.5% remaining clearly spans 2 half-lives.

Answer 1

The Correct Option is C

Concept: Radioactive decay follows a first-order kinetics process where the half-life (\(T_{1/2}\)) is the time required for half of the initial quantity of a substance to decay.
• Amount Remaining: After \(n\) half-lives, the remaining amount \(N\) is \(N = N_0 \left( \frac{1}{2} \right)^n\).
• Decay Percentages: Decay of 50% means 50% remains (\(N_1 = 0.5N_0\)). Decay of 87.5% means 12.5% remains (\(N_2 = 0.125N_0\)).

Step 1: Determine the number of half-lives for each state.
- For 50% decay (50% remaining): \[ 0.5 = (1/2)^1 \implies n_1 = 1 \text{ half-life} \] - For 87.5% decay (12.5% remaining): \[ 0.125 = \frac{1}{8} = (1/2)^3 \implies n_2 = 3 \text{ half-lives} \]

Step 2: Calculate the time difference.
The time interval between these two states is the difference in the number of half-lives: \[ \Delta n = n_2 - n_1 = 3 - 1 = 2 \text{ half-lives} \] Since \(T_{1/2} = 20\) minutes: \[ \text{Time taken} = 2 \times 20 = 40 \text{ minutes} \]