Question

A radioactive material is reduced to \( \frac{1}{8} \) of its original amount in 3 days. If \( 8 \times 10^{-3} \) kg of the material is left after 5 days, what was the initial amount of the material?

• A. \( 700 \) gm
• B. \( 900 \) gm
• C. \( 475 \) gm
• D. \( 256 \) gm

Hint:

For radioactive decay problems: - Identify the half-life first. - Use the exponential decay formula: \[ N = N_0 \times \left(\frac{1}{2}\right)^{t/T} \] - Ensure unit conversions (kg to g) for final answers.

Answer 1

The Correct Option is D

Step 1: Understanding radioactive decay. The decay of a radioactive material follows the exponential decay law: \[ N = N_0 \times \left(\frac{1}{2}\right)^{t/T} \] where: - \( N_0 \) = Initial amount of substance - \( N \) = Remaining amount after time \( t \) - \( T \) = Half-life - \( t \) = Given time 
Step 2: Finding the half-life. Given that the material is reduced to \( \frac{1}{8} \) of its original value in 3 days, we set up: \[ \frac{N_0}{8} = N_0 \times \left(\frac{1}{2}\right)^{3/T} \] Since \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), we compare powers: \[ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{3/T} \] Thus, \( T = 1 \) day (half-life is 1 day). 
Step 3: Finding \( N_0 \). After 5 days, the amount left is \( 8 \times 10^{-3} \) kg. Using the decay formula: \[ N = N_0 \times \left(\frac{1}{2}\right)^{5/1} \] \[ 8 \times 10^{-3} = N_0 \times \frac{1}{32} \] Solving for \( N_0 \): \[ N_0 = 8 \times 10^{-3} \times 32 = 0.256 { kg} = 256 { g} \] Final Answer: \[ \boxed{256 { gm}} \]