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Chapter 29: Problem 67

\(_{52}^{106} \mathrm{Te}\) is radioactive; it \(\alpha\) decays to $^{102} \mathrm{s}_{0} \mathrm{Sn} .\( " \)_{50}^{102} \mathrm{Sn}$ is itself radioactive and has a half-life of 4.5 s. At \(t=0,\) a sample contains 4.00 mol of \(^{106}\) Te and 1.50 mol of \(^{102}\) so Sn. At \(t=\) \(25 \mu \mathrm{s},\) the sample contains \(3.00 \mathrm{mol}\) of \(^{106} \mathrm{Te}\) and $2.50 \mathrm{mol}\( of \)^{102} \mathrm{s} 0$ Sn. How much 102 50 Sn will there be at \(t=50 \mu \mathrm{s} ?\)

Short Answer

Expert verified
Answer: Approximately 3.42 moles

Step by step solution

01

Find the decay constant of \(_{50}^{102}\mathrm{Sn}\)

Using the half-life formula, we can determine the decay constant \(\lambda_{Sn}\) for \(_{50}^{102}\mathrm{Sn}\): \[T_{1/2} = \frac{ln(2)}{\lambda}\] Given half-life \(T_{1/2} = 4.5\mathrm{s}\), solve for \(\lambda_{Sn}\): \[\lambda_{Sn} = \frac{ln(2)}{4.5}\] Now calculate \(\lambda_{Sn}\): \[\lambda_{Sn} \approx 0.154\,\mathrm{s}^{-1}\]
02

Write the equation for moles of \(_{50}^{102}\mathrm{Sn}\) and \(_{52}^{106}\mathrm{Te}\)

At time \(t\), the equation for the moles of \(_{50}^{102}\mathrm{Sn}\) is: \[n_{Sn}(t) = n_{Sn0}e^{-\lambda_{Sn}t}\] And the equation for moles of \(_{52}^{106}\mathrm{Te}\) is: \[n_{Te}(t) = n_{Te0}e^{-\lambda_{Te}t}\] We are given 2 sets of values: 1. At \(t = 0\): \[n_{Te0} = 4.00\,mol\] \[n_{Sn0} = 1.50\,mol\] 2. At \(t = 25\,\mu s\): \[n_{Te}(25\,\mu s) = 3.00\,mol\] \[n_{Sn}(25\,\mu s) = 2.50\,mol\]
03

Find the decay constant of \(_{52}^{106}\mathrm{Te}\)

Use the given information to find \(\lambda_{Te}\): \[3.00 = 4.00 e^{-\lambda_{Te}(25\,\mu s)}\] Solve for \(\lambda_{Te}\): \[\lambda_{Te} = \frac{ln(4/3)}{25\,\mu s}\] Now calculate \(\lambda_{Te}\): \[\lambda_{Te} \approx 1.849 \times 10^{-3}\,\mathrm{s}^{-1}\]
04

Calculate the amount of \(_{50}^{102}\mathrm{Sn}\) at \(t = 50\, \mu s\)

To calculate the amount of \(_{50}^{102}\mathrm{Sn}\) at \(t = 50\, \mu s\), use the following equations: \[n_{Te}(50\,\mu s) = 4.00 e^{-\lambda_{Te}(50\,\mu s)}\] \[n_{Sn}(50\,\mu s) = \left[n_{Sn0} + \left(n_{Te0} - n_{Te}(50\,\mu s)\right)\right]e^{-\lambda_{Sn}(50\,\mu s)}\] First, calculate \(n_{Te}(50\,\mu s)\): \[n_{Te}(50\,\mu s) \approx 2.08\,mol\] Now, calculate \(n_{Sn}(50\,\mu s)\): \[n_{Sn}(50\,\mu s) = \left[1.50 + \left(4.00 - 2.08\right)\right]e^{-0.154 \times 50 \times 10^{-6}}\] Finally, calculate the value: \[n_{Sn}(50\,\mu s) \approx 3.42\,mol\] So, at \(t = 50\, \mu s\), there will be approximately 3.42 moles of \(_{50}^{102}\mathrm{Sn}\).

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