/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 In \(1992,\) Swiss police arrest... [FREE SOLUTION] | 91Ó°ÊÓ

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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.)

Short Answer

Expert verified
The activity of each \(1.0 \mathrm{~g}\) \(^{137} \mathrm{Cs}\) sample is approximately 3.20 trillion becquerels or 86.5 millicuries.

Step by step solution

01

Convert Half-life from Years to Seconds

The half-life of \({ }^{137} \mathrm{Cs}\) is given as 30.2 years. First, we need to convert this time period into seconds to use in activity calculations: \\[\text{1 year} = 365 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 31,536,000 \text{ seconds}\] \Now calculate half-life in seconds: \\[t_{1/2} = 30.2 \times 31,536,000 = 9.52 \times 10^8 \text{ seconds}\]
02

Calculate Decay Constant (λ)

Using the formula \(\lambda = \frac{\ln(2)}{t_{1/2}}\), find the decay constant. \\[\lambda = \frac{0.693}{9.52 \times 10^8} \approx 7.28 \times 10^{-10} \text{ s}^{-1}\]
03

Determine the Number of Atoms in the Sample

To find the number of atoms in \(1.0 \text{ g}\) of \(^{137}\mathrm{Cs}\), calculate as follows: \The molar mass of \(^{137}\mathrm{Cs}\) is approximately 137 g/mol. Hence, \\[\text{Number of moles} = \frac{1.0 \text{ g}}{137 \text{ g/mol}} = 7.3 \times 10^{-3} \text{ moles}\] \Using Avogadro's number (\(6.022 \times 10^{23} \text{ atoms/mol}\)), calculate: \\[N = 7.3 \times 10^{-3} \times 6.022 \times 10^{23} = 4.40 \times 10^{21} \text{ atoms}\]
04

Calculate Activity in Becquerels

Activity \(A\) in becquerels is given by \(A = \lambda N\). \\[A = 7.28 \times 10^{-10} \times 4.40 \times 10^{21} \A \approx 3.20 \times 10^{12} \text{ Becquerels}\]
05

Convert Activity to Curies

To convert from becquerels to curies: \(1 \text{ Curie} = 3.7 \times 10^{10} \text{ Bq}\). \\[A = \frac{3.20 \times 10^{12} \text{ Bq}}{3.7 \times 10^{10} \text{ Bq/Curie}} \A \approx 86.5 \text{ mCi} \\] \Thus, the activity of each sample is approximately 86.5 millicuries.

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Key Concepts

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

Half-Life Calculation
Radioactive decay is a spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. A key concept in understanding radioactive decay is the **half-life**, which is the time required for half of the radioactive atoms in a sample to decay. It's a constant value specific to each radioactive isotope.
To calculate the half-life of a substance in different time units, it's essential to understand conversion factors. For instance, the given half-life of \(^{137}\text{Cs}\) is 30.2 years. If you want to convert this into seconds, use:
  • 1 year = 31,536,000 seconds
  • Therefore, \(t_{1/2} = 30.2 \times 31,536,000 = 9.52 \times 10^8\) seconds
Learning to perform these conversions is crucial for solving many radioactive decay problems. This allows you to apply the half-life to different operational contexts, such as medical applications or environmental studies.
Becquerel and Curie Conversion
In radioactive measurements, the **activity** of a radioactive sample is often expressed in two units: becquerels (Bq) and curies (Ci). These units indicate how many atoms in the sample decay in a given period.
  • 1 Becquerel (Bq) is defined as one decay per second.
  • 1 Curie (Ci) is equivalent to \(3.7 \times 10^{10}\) becquerels.
Knowing how to convert between these two units is essential when comparing activities.
For instance, if we start with the activity in becquerels, calculated as \(3.20 \times 10^{12}\) Bq for a \(1\, \text{g}\) sample of \(^{137}\text{Cs}\), converting to curies gives:\[A = \frac{3.20 \times 10^{12}\, \text{Bq}}{3.7 \times 10^{10}\, \text{Bq/Ci}} \approx 86.5\, \text{mCi}\]Understanding this conversion helps you to interpret and communicate radioactive levels in a universally accepted way. This is particularly useful in fields like nuclear medicine, where precise measurements are vital for patient safety and treatment efficacy.
Decay Constant
The **decay constant** \(\lambda\) is a probability measure of how quickly a radioactive substance undergoes decay. It's directly related to the half-life and provides a mathematical approach to calculate the activity of a radioactive sample.
The formula is \(\lambda = \frac{\ln(2)}{t_{1/2}}\). This equation arises from the exponential nature of radioactive decay. Here, \(\ln(2) \approx 0.693\) is a constant since radioactive decay follows a first-order kinetic process. This means the process speed is proportional to the number of undecayed nuclei.
For \(^{137}\text{Cs}\), where \(t_{1/2} = 9.52 \times 10^8\) seconds:\[\lambda = \frac{0.693}{9.52 \times 10^8} \approx 7.28 \times 10^{-10} \text{ s}^{-1}\]Having the decay constant allows us to calculate the activity \(A\) of the sample using the formula \(A = \lambda N\), where \(N\) is the number of undecayed atoms. This concept is fundamental in calculations involving decay rates and in predicting how long a substance will remain hazardous.

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Most popular questions from this chapter

What is the binding energy per nucleon of the rutherfordium isotope \({ }_{104}^{259} \mathrm{Rf}\) ? Here are some atomic masses and the neutron mass. $$\begin{array}{lr}\frac{259}{104} \mathrm{Rf} & 259.10563 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

The radionuclide \({ }^{11} \mathrm{C}\) decays according to $${ }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \mathrm{~min}$$ 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.)

Radioactive element \(A A\) can decay to either element \(B B\) or element \(C C\). The decay depends on chance, but the ratio of the resulting number of \(B B\) atoms to the resulting number of \(C C\) atoms is always \(2 / 1 .\) The decay has a half-life of 8.00 days. We start with a sample of pure \(A A .\) How long must we wait until the number of \(C C\) atoms is 1.50 times the number of \(A A\) atoms?

How much energy is released when a \({ }^{238} \mathrm{U}\) nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are $$ \begin{array}{llll} { }^{238} \mathrm{U} & 238.05079 \mathrm{u} & { }^{234} \mathrm{Th} & 234.04363 \mathrm{u} \\ { }^{237} \mathrm{U} & 237.04873 \mathrm{u} & { }^{4} \mathrm{He} & 4.00260 \mathrm{u} \\ { }^{236} \mathrm{~Pa} & 236.04891 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ { }^{235} \mathrm{~Pa} & 235.04544 \mathrm{u} & \mathrm{n} & 1.00866 \mathrm{u} \end{array} $$

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 \(\mathrm{RBE}\) 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.

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