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Chapter 29: Problem 38

The activity of a sample containing radioactive \(^{108} \mathrm{Ag}\) is $6.4 \times 10^{4}\( Bq. Exactly 12 min later, the activity is \)2.0 \times 10^{3} \mathrm{Bq} .\( Calculate the half-life of \)^{108} \mathrm{Ag} .$

Short Answer

Expert verified
Answer: The half-life of radioactive \(^{108}\mathrm{Ag}\) is approximately 2.4 minutes.

Step by step solution

01

Understand the radioactive decay law

The activity of a radioactive sample, A, is given by the radioactive decay law: \(A(t) = A_0 e^{-\lambda t}\) where \(A_0\) is the initial activity, \(t\) is the elapsed time, and \(\lambda\) is the decay constant.
02

Derive the half-life formula

To derive the half-life formula, we can use the definition of half-life (\(T_{1/2}\)), which is the time it takes for the activity to reduce to half of its initial value: \(\frac{1}{2}A_0 = A_0 e^{-\lambda T_{1/2}}\) Divide both sides by \(A_0\): \(\frac{1}{2} = e^{-\lambda T_{1/2}}\) Take the natural logarithm of both sides: \(-\ln{2} = -\lambda T_{1/2}\) Now solve for \(T_{1/2}\): \(T_{1/2} = \frac{\ln{2}}{\lambda}\) We now have a formula to calculate the half-life of the radioactive sample.
03

Calculate the decay constant \(\lambda\)

Using the data given in the problem, we can set up an equation to find the decay constant \(\lambda\). \(\begin{aligned}A(t) &= A_0 e^{-\lambda t}\\ 2.0 \times 10^3\, \mathrm{Bq} &= (6.4 \times 10^4\, \mathrm{Bq}) e^{-\lambda(12\,\text{min})}\end{aligned}\) Divide both sides by \(6.4 \times 10^4\, \mathrm{Bq}\): \(0.03125 = e^{-\lambda(12\,\text{min})}\) Take the natural logarithm of both sides: \(-\ln{32} = -\lambda(12\,\text{min})\) Divide by \(-12\,\text{min}\): \(\lambda = \frac{\ln{32}}{12\,\text{min}}\)
04

Calculate the half-life \(T_{1/2}\)

Using the derived half-life formula and the decay constant from Step 3, we can now calculate the half-life of the radioactive \(^{108}\mathrm{Ag}\): \(T_{1/2} = \frac{\ln{2}}{\lambda} = \frac{\ln{2}}{\frac{\ln{32}}{12\,\text{min}}}\) Multiply by \(12\,\text{min}\): \(T_{1/2} = \frac{12 \times \ln{2}}{\ln{32}}\) Calculate the value: \(T_{1/2} \approx 2.4\,\text{min}\) The half-life of radioactive \(^{108}\mathrm{Ag}\) is approximately 2.4 minutes.

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