Question

A radioactive material of half-life 2.5 hours emits radiation that is 32 times the safe maximum level. The time (in hours) after which the material can be handled safely is

• A. 10
• B. 25
• C. 5
• D. 12.5

Hint:

For radioactive decay problems where the reduction factor is a power of 2 (like 2, 4, 8, 16, 32, 64...), it's quickest to determine the number of half-lives directly. Since \(32 = 2^5\), it takes 5 half-lives for the activity to decrease by a factor of 32.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept: Radioactive decay follows the law \( A = A_0 (1/2)^n \), where \( A \) is the current activity, \( A_0 \) is the initial activity, and \( n \) is the number of half-lives. Here, the initial radiation level is 32 times the safe level. Let the safe level be \( S \). So, \( A_0 = 32S \) and we need to find the time when the activity \( A \) becomes \( S \).
Step 2: Calculate the Number of Half-Lives: Using the relation: \[ \frac{A}{A_0} = \left(\frac{1}{2}\right)^n \] \[ \frac{S}{32S} = \left(\frac{1}{2}\right)^n \] \[ \frac{1}{32} = \left(\frac{1}{2}\right)^n \] Since \( 32 = 2^5 \), we have: \[ \left(\frac{1}{2}\right)^5 = \left(\frac{1}{2}\right)^n \] \[ n = 5 \]
Step 3: Calculate the Time: The total time \( t \) is given by: \[ t = n \times T_{1/2} \] Given \( T_{1/2} = 2.5 \) hours. \[ t = 5 \times 2.5 = 12.5 \text{ hours} \]