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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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Most popular questions from this chapter

The power supply for a pacemaker is a small amount of radioactive $^{238} \mathrm{Pu} .\( This nuclide decays by \)\alpha$ decay with a half-life of 86 yr. The pacemaker is typically replaced every 10.0 yr. (a) By what percentage does the activity of the \(^{238}\) Pu source decrease in 10 yr? (b) The energy of the \(\alpha\) particles emitted is 5.6 MeV. Assume an efficiency of \(100 \%\) -all of the \(\alpha\) -particle energy is used to run the pacemaker. If the pacemaker starts with \(1.0 \mathrm{mg}\) of \(^{238} \mathrm{Pu}\) what is the power output initially and after 10.0 yr?
Calculate the activity of \(1.0 \mathrm{g}\) of radium- 226 in Ci.
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?
Radium- 226 decays as ${ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} .\( If the \){ }_{88}^{226}$ Ra nucleus is at rest before the decay and the \({ }_{86}^{222} \mathrm{Rn}\) nucleus is in its ground state, estimate the kinetic energy of the \(\alpha\) particle. (Assume that the \({ }_{86}^{222} \mathrm{Rn}\) nucleus takes away an insignificant fraction of the kinetic energy.)
Write the symbol (in the form \(_{Z}^{A} \mathrm{X}\) ) for the nuclide that has 78 neutrons and 53 protons.
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