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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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Most popular questions from this chapter

A certain radioactive nuclide has a half-life of \(200.0 \mathrm{s}\) A sample containing just this one radioactive nuclide has an initial activity of \(80,000.0 \mathrm{s}^{-1} .\) (a) What is the activity 600.0 s later? (b) How many nuclei were there initially? (c) What is the probability per second that any one of the nuclei decays?
Show that the spontaneous \(\alpha\) decay of \(^{19} \mathrm{O}\) is not possible.
An \(\alpha\) particle produced in radioactive \(\alpha\) decay has a kinetic energy of typically about 6 MeV. When an \(\alpha\) particle passes through matter (e.g., biological tissue), it makes ionizing collisions with molecules, giving up some of its kinetic energy to supply the binding energy of the electron that is removed. If a typical ionization energy for a molecule in the body is around \(20 \mathrm{eV}\) roughly how many molecules can the alpha particle ionize before coming to rest?
Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by electron capture.
The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
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