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Chapter 29: Problem 34

The half-life of I-131 is 8.0 days. A sample containing I- 131 has an activity of \(6.4 \times 10^{8}\) Bq. How many days later will the sample have an activity of \(2.5 \times 10^{6} \mathrm{Bq} ?\) (Wills tutorial: radioactive decay)

Short Answer

Expert verified
Answer: It takes approximately 29.6 days for the I-131 sample's activity to reduce from \(6.4 \times 10^{8} \mathrm{Bq}\) to \(2.5 \times 10^{6} \mathrm{Bq}\).

Step by step solution

01

Calculate the decay constant \(\lambda\)

Using the half-life formula: \(\lambda = \frac{ln(2)}{t_{1/2}}\), we can find the decay constant. \(t_{1/2} = 8.0 \, \text{days}\) (given) \(\lambda = \frac{ln(2)}{8.0}\)
02

Set up the radioactive decay equation

Now that we have the decay constant, λ, we can use the radioactive decay formula to find the time t. \(A = A_0 e^{-\lambda t}\) We are given: - \(A = 2.5 \times 10^{6} \, \text{Bq}\) - \(A_0 = 6.4 \times 10^{8} \, \text{Bq}\) We will now plug these values into the equation: \(2.5 \times 10^{6} = 6.4 \times 10^{8} e^{-\lambda t}\)
03

Solve for t

To solve for \(t\), we first isolate \(e^{-\lambda t}\) by dividing both sides by \(6.4 \times 10^{8}\): \(e^{-\lambda t} = \frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\) Now, we will take the natural logarithm of both sides to isolate the \(-\lambda t\) term: \(-\lambda t = ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)\) Finally, we divide by \(-\lambda\) to solve for \(t\). Since \(\lambda = \frac{ln(2)}{8}\), we have: \(t = \frac{ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)}{-\frac{ln(2)}{8}}\)
04

Calculate the time

Now we will calculate the value of \(t\) using the values we derived: \(t = \frac{ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)}{-\frac{ln(2)}{8}} = \boxed{29.6} \, \text{days}\) (approximately) So, it would take around 29.6 days for the I-131 sample's activity to reduce from \(6.4 \times 10^{8} \mathrm{Bq}\) to \(2.5 \times 10^{6} \mathrm{Bq}\).

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