Question

A radioactive element \(A\) converts into another stable element \(B\). Half-life of \(A\) is \(1.5\,\text{hrs}\). After time \(t\), the ratio of atoms of \(A\) and \(B\) is found to be \(1:8\), then \(t\) in hours is

• A. \(6\)
• B. \(8\)
• C. Between \(3\) to \(4.5\)
• D. Between \(4.5\) to \(6\)

Hint:

If a radioactive element \(A\) decays into stable \(B\), then \[ B=N_0-N. \] Use the given ratio \(A:B\) to first find the remaining fraction of \(A\), then compare it with powers of \(\frac12\).

Answer 1

The Correct Option is D

Step 1: Set up decay with stable daughter.
Initially N₀ atoms of A. At time t, N_A = N, N_B = N₀ – N. Given N_A : N_B = 1 : 8 → N/(N₀–N) = 1/8 → N = N₀/9.

Step 2: Apply exponential decay law.
N = N₀(1/2)^(t/T). N₀/9 = N₀(1/2)^(t/1.5) → 1/9 = (1/2)^(t/1.5).

Step 3: Bracket the time using half-life multiples.
After 3T (4.5 hrs): N = N₀/8. After 4T (6 hrs): N = N₀/16. N₀/9 lies between, so t ∈ (4.5, 6) hrs.

Step 4: Final Answer:
Between 4.5 to 6 hours.