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Multiple Choice
If 60% of a radioactive isotope sample decays in 24 hours, what is its half-life?
A
10 hours
B
15 hours
C
20 hours
D
24 hours
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Verified step by step guidance
1
Understand that the decay of a radioactive isotope follows first-order kinetics, which can be described by the equation: \( N_t = N_0 e^{-kt} \), where \( N_t \) is the remaining quantity of the substance, \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is time.
Since 60% of the sample decays, 40% remains. Therefore, \( N_t = 0.4N_0 \). Substitute this into the decay equation: \( 0.4N_0 = N_0 e^{-kt} \).
Simplify the equation by dividing both sides by \( N_0 \), resulting in \( 0.4 = e^{-kt} \).
Take the natural logarithm of both sides to solve for \( k \): \( \ln(0.4) = -kt \). Rearrange to find \( k = -\frac{\ln(0.4)}{t} \).
Use the relationship between the decay constant \( k \) and half-life \( t_{1/2} \), which is \( t_{1/2} = \frac{\ln(2)}{k} \). Substitute \( k \) from the previous step to find the half-life.