Question

2 kg of a pure radioactive substance after a passage of \(N\) years is left with only 125 g of pure substance. The half-life of the radioactive substance is 12.5 y. Find the value of \(N\).

• A. 50 y
• B. 37.5 y
• C. 25 y
• D. 100 y

Hint:

Use \(N_t = N_0 (1/2)^{t/T_{1/2}}\) for decay calculations. Express remaining fraction as a power of 2 to solve for elapsed time.

Answer 1

The Correct Option is A

Step 1: Decay law.
N_t = N₀ (1/2)^(t/T₁/₂), with N₀ = 2 kg, N_t = 0.125 kg, T₁/₂ = 12.5 years.

Step 2: Substituting values.
0.125 = 2 (1/2)^(t/12.5).

Step 3: Simplifying.
0.125/2 = (1/2)^(t/12.5) → 1/16 = (1/2)^(t/12.5).

Step 4: Equating exponents.
1/16 = (1/2)⁴ → (1/2)^(t/12.5) = (1/2)⁴ → t/12.5 = 4 → t = 50 years.

Step 5: Conclusion.
The decay from 2 kg to 125 g requires 50 years.