Multiple Choice
Estimate the half-life of the isotope from the graph below.
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21. Nuclear Chemistry
Rate of Radioactive Decay
Multiple Choice
For the radioactive decay of lead-202 the decay constant is 1.32 × 10-5 yr-1. How long will it take in hours to decrease to 53% of its initial amount?
A
5.72 × 104 hrs
B
5.01 × 109 hrs
C
4.81 × 104 hrs
D
4.21 × 108 hrs
Verified step by step guidance1
Identify the given data: the decay constant \(\lambda = 1.32 \times 10^{-5} \text{ yr}^{-1}\) and the fraction remaining \(\frac{N}{N_0} = 0.53\) (53% of the initial amount).
Recall the radioactive decay formula relating the remaining amount to time:
\(\frac{N}{N_0} = e^{-\lambda t}\)
Rearrange the formula to solve for time \(t\):
\(t = -\frac{1}{\lambda} \ln\left(\frac{N}{N_0}\right)\)
Substitute the known values into the equation:
\(t = -\frac{1}{1.32 \times 10^{-5} \text{ yr}^{-1}} \ln(0.53)\)
Calculate \(t\) in years first, then convert the time from years to hours by multiplying by the number of hours in one year:
\(\text{hours} = t \times 365 \times 24\)
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