If the half-life of an element is 69.3 hours, then how much of its decay in 10th to 11th hours? Initial activity = 50µCi.

Correct option is A.1

The decay constant λ is given by:
λ = ln(2) / t½
where t½ is the half-life.

Given t½ = 69.3 hours
λ = ln(2) / 69.3 = 0.01 hour⁻¹

The activity A(t) at time t is given by:
A(t) = A₀ * e^(-λt)
where A₀ is the initial activity.

Given A₀ = 50 µCi

Activity at t = 10 hours:
A(10) = 50 * e^(-0.01 * 10) = 50 * e^(-0.1) ≈ 45.2 µCi

Activity at t = 11 hours:
A(11) = 50 * e^(-0.01 * 11) = 50 * e^(-0.11) ≈ 44.7 µCi

Decay in 10th to 11th hour = A(10) - A(11) ≈ 45.2 µCi - 44.7 µCi ≈ 0.5 µCi

Since the options are integers, we can approximate the decay as 1 µCi. Therefore, the closest option is A (1).