Multiple Choice
Estimate the half-life of the isotope from the graph below.
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21. Nuclear Chemistry
Rate of Radioactive Decay
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For the radioactive decay of lead-202 the decay constant is 1.32 x 10-5 yr-1. How long will it take in hours to decrease to 53% of its initial amount?
A
5.72 x 104 hrs
B
5.01 x 109 hrs
C
4.81 x 104 hrs
D
4.21 x 108 hrs
Verified step by step guidance1
Identify the formula for radioactive decay: \( N(t) = N_0 e^{-\lambda t} \), where \( N(t) \) is the remaining quantity, \( N_0 \) is the initial quantity, \( \lambda \) is the decay constant, and \( t \) is time.
Since we want to find the time \( t \) when the quantity decreases to 53% of its initial amount, set \( N(t) = 0.53 N_0 \).
Substitute \( N(t) = 0.53 N_0 \) into the decay formula: \( 0.53 N_0 = N_0 e^{-\lambda t} \).
Cancel \( N_0 \) from both sides to get: \( 0.53 = e^{-\lambda t} \).
Take the natural logarithm of both sides to solve for \( t \): \( \ln(0.53) = -\lambda t \). Rearrange to find \( t = \frac{\ln(0.53)}{-\lambda} \). Convert \( t \) from years to hours by multiplying by the number of hours in a year (365.25 days/year * 24 hours/day).
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