What is the decay constant for a radioactive isotope which decreases to 34% of its original value in 2.48 hours? A)0.137 hr^-1 B)0.168 hr^-1 C)0.435 hr^-1 D)2.30 hr^-1

Question

Quizplus · Accepted Answer

We can use the formula for radioactive decay: 
N = N0 * e^(-λt) 
where N = final amount (34% of original), N0 = initial amount (100%), λ = decay constant, and t = time (2.48 hours).

Substituting the given values: 
0.34 = 1 * e^(-λ*2.48)

Solving for λ: 
λ = ln(0.34)/(-2.48) 
λ ≈ 0.435 hr^-1

Therefore, the decay constant is 0.435 hr^-1, which is closest to choice C (0.435 hr^-1).

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