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Chapter 42: Problem 8

What is the activity of a \(10 \mathrm{ng}\) sample of \({ }^{92} \mathrm{Kr}\), which has a halflife of \(1.84 \mathrm{~s}\) ?

Short Answer

Expert verified
The activity of the sample is calculated by multiplying the decay constant by the number of nuclei.

Step by step solution

01

Understand the Given Information

We are given a sample of \({ }^{92} \mathrm{Kr}\) with a mass of \(10 \mathrm{ng}\) and a half-life of \(1.84 \mathrm{s}\). Our goal is to find the activity of this sample, typically expressed in units of Becquerels (Bq), which is the unit of decay events per second.
02

Use the Decay Constant Formula

The decay constant, \(\lambda\), is related to the half-life, \(T_{1/2}\), by the formula \(\lambda = \frac{\ln 2}{T_{1/2}}\). Substituting \(T_{1/2} = 1.84 \mathrm{s}\), we compute:\[\lambda = \frac{\ln 2}{1.84}\]
03

Find the Number of Nuclei, \(N\)

Use the mass of the sample and Avogadro's number to find the number of \(^{92} \mathrm{Kr}\) nuclei. First, convert the mass from grams to moles: 1 mole of \(^{92} \mathrm{Kr}\) has a mass of \(92 \mathrm{g/mol}\). Thus, \(10 \mathrm{ng} = 10\times10^{-9}\mathrm{g}\), and the number of moles is \[\text{moles} = \frac{10 \times 10^{-9}}{92}\].Then, find \(N = \text{moles} \times \text{Avogadro's number, } 6.022 \times 10^{23}\mathrm{mol}^{-1}\).
04

Calculate the Activity, \(A\)

Activity is expressed as \(A = \lambda N\). Using \(\lambda\) from Step 2 and \(N\) from Step 3, compute \[ A = (\lambda \times N) \] to find the activity in Becquerels.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Decay Constant
In the context of radioactivity, the decay constant ( \(\lambda\)) is a fundamental parameter that signifies how quickly a radioactive substance decays. It's inherently linked to the half-life of a radioactive element.
For example, the decay constant is defined by the relation \(\lambda = \frac{\ln 2}{T_{1/2}}\), where \(T_{1/2}\) represents the half-life.
This formula shows that the decay constant and half-life are inversely related.
  • If the half-life is short, the decay constant is large, indicating rapid decay.
  • Conversely, a long half-life indicates a small decay constant and a slower decay process.
Understanding \(\lambda\) helps in calculating how many decay events occur over time, providing insight into the nature of the sample's radioactivity.
Half-life
The half-life of a radioactive isotope is the time it takes for half of the sample's nuclei to decay. It's a crucial aspect that helps predict how long a radioactive sample remains active.
The half-life is significant because:
  • It provides a fixed measure allowing comparison between different isotopes.
  • It is key in determining the time over which a sample remains hazardous.
In our exercise, the half-life of \(^{92}\mathrm{Kr}\) is \(1.84\ \text{s}\). This short half-life suggests that the isotope decays quickly, proving essential for calculations related to the sample鈥檚 radioactivity.
Becquerel (Bq)
The Becquerel (Bq) is the standard unit used in measuring the activity of a radioactive material, named after Henri Becquerel, who, along with Marie and Pierre Curie, discovered radioactivity.
One Bq is equivalent to one decay event per second.
  • This straightforward measurement allows scientists to quantify the activity level within a sample accurately.
  • Comparing Bq between different samples can indicate how intensely they exhibit radioactive decay.
Calculating the activity in Bq helps researchers and safety personnel assess both the potential exposure risk and the presence of radioactive material in environments or surfaces.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\ \text{mol}^{-1}\), is a fundamental constant representing the number of atoms, ions, or molecules in one mole of a substance.
It's critical in calculations involving chemical quantities and radioactive decay analysis.
In dealing with radioactivity and our specific exercise:
  • Avogadro's number enables conversion from the macroscopic world of grams to the microscopic level of individual nuclei.
  • By using this number, we can assess the total number of particles available for decay, which directly influences the computed activity.
This bridge between the atomic and macroscopic realms is indispensable for thorough scientific investigation and understanding.

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Most popular questions from this chapter

The nuclide \({ }^{198} \mathrm{Au}\), with a half-life of \(2.70 \mathrm{~d}\), is used in cancer therapy. What mass of this nuclide is required to produce an activity of \(250 \mathrm{Ci}\) ?

An \(85 \mathrm{~kg}\) worker at a breeder reactor plant accidentally ingests \(2.5 \mathrm{mg}\) of \({ }^{239} \mathrm{Pu}\) dust. This isotope has a half- life of \(24100 \mathrm{y}\), decaying by alpha decay. The energy of the emitted alpha particles is \(5.2 \mathrm{MeV}\), with an RBE factor of 13 . Assume that the plutonium resides in the worker's body for \(12 \mathrm{~h}\) (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that \(95 \%\) of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the \(12 \mathrm{~h}\), (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

In 1992, Swiss police arrested two men who were attempting to smuggle osmium out of Eastern Europe for a clandestine sale. However, by error, the smugglers had picked up \({ }^{137} \mathrm{Cs}\). Reportedly, each smuggler was carrying a \(1.0 \mathrm{~g}\) sample of \({ }^{137} \mathrm{Cs}\) in a pocket! In (a) bequerels and (b) curies, what was the activity of each sample? The isotope \({ }^{137} \mathrm{Cs}\) has a half-life of \(30.2 \mathrm{y}\). (The activities of radioisotopes commonly used in hospitals range up to a few millicuries.)

The radionuclide \({ }^{11} \mathrm{C}\) decays according to $$ { }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \min $$ The maximum energy of the emitted positrons is \(0.960 \mathrm{MeV}\). (a) Show that the disintegration energy \(Q\) for this process is given by $$ Q=\left(m_{\mathrm{C}}-m_{\mathrm{B}}-2 m_{\mathrm{e}}\right) c^{2} $$ where \(m_{\mathrm{C}}\) and \(m_{\mathrm{B}}\) are the atomic masses of \({ }^{11} \mathrm{C}\) and \({ }^{11} \mathrm{~B}\), respectively, and \(m_{e}\) is the mass of a positron. (b) Given the mass values \(m_{\mathrm{C}}=11.011434 \mathrm{u}, m_{\mathrm{B}}=11.009305 \mathrm{u}\), and \(m_{\mathrm{e}}=0.0005486 \mathrm{u}\), calculate \(Q\) and compare it with the maximum energy of the emitted positron given above. (Hint: Let \(\mathbf{m}_{\mathrm{C}}\) and \(\mathbf{m}_{\mathrm{B}}\) be the nuclear masses and then add in enough electrons to use the atomic masses.)

Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) \({ }^{235} \mathrm{U} \rightarrow{ }^{232} \mathrm{Th}+{ }^{3} \mathrm{He}, \quad\) (b) \({ }^{235} \mathrm{U} \rightarrow{ }^{231} \mathrm{Th}+{ }^{4} \mathrm{He}\), (c) \({ }^{235} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+{ }^{5} \mathrm{He}\). The needed atomic masses are \(\begin{array}{llll}{ }^{232} \mathrm{Th} & 232.0381 \mathrm{u} & { }^{3} \mathrm{He} & 3.0160 \mathrm{u} \\ { }^{21} \mathrm{Th} & 231.0363 \mathrm{u} & { }^{4} \mathrm{He} & 4.0026 \mathrm{u} \\ { }^{230} \mathrm{Th} & 230.0331 \mathrm{u} & { }^{5} \mathrm{He} & 5.0122 \mathrm{u} \\ { }^{235} \mathrm{U} & 235.0429 \mathrm{u} & & \end{array}\)
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