/*! 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 53 The cesium isotope \({ }^{137} \... [FREE SOLUTION] | 91影视

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Chapter 42: Problem 53

The cesium isotope \({ }^{137} \mathrm{Cs}\) is present in the fallout from aboveground detonations of nuclear bombs. Because it decays with a slow \((30.2 \mathrm{y})\) half-life into \({ }^{137} \mathrm{Ba},\) releasing considerable energy in the process, it is of environmental concern. The atomic masses of the Cs and \(\mathrm{Ba}\) are 136.9071 and \(136.9058 \mathrm{u},\) respectively; calculate the total energy released in such a decay.

Short Answer

Expert verified
The total energy released is 1.213 MeV.

Step by step solution

01

Determine the Mass Defect

The mass defect, \( \Delta m \), is the difference in mass between the initial isotope \( \left( {}^{137} \mathrm{Cs} \right) \) and the final isotope \( \left( {}^{137} \mathrm{Ba} \right) \). It can be calculated as \( \Delta m = m_\text{Cs} - m_\text{Ba} \), where \( m_\text{Cs} = 136.9071 \, \text{u} \) and \( m_\text{Ba} = 136.9058 \, \text{u} \). Substituting these values gives \( \Delta m = 136.9071 - 136.9058 = 0.0013 \, \text{u} \).
02

Convert Mass Defect to Energy

Using the mass-energy equivalence principle, \( E = \Delta m c^2 \), where \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). First, convert \( \Delta m \) from atomic mass units to kilograms using \( 1 \, \text{u} = 1.660539 \times 10^{-27} \text{ kg} \). Thus, \( \Delta m = 0.0013 \, \text{u} \times 1.660539 \times 10^{-27} \text{ kg/u} = 2.1587 \times 10^{-30} \text{ kg} \). Then, \( E = 2.1587 \times 10^{-30} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \).
03

Calculate the Energy Released

Calculate the energy: \[ E = 2.1587 \times 10^{-30} \text{ kg} \times 9 \times 10^{16} \text{ m}^2/\text{s}^2 = 1.94283 \times 10^{-13} \text{ Joules} \]. This is the total energy released during the decay of one atom of \( {}^{137}\text{Cs} \) into \( {}^{137}\text{Ba} \).
04

Convert Energy to MeV

To convert the energy from Joules to mega-electronvolts (MeV), use the conversion factor \( 1 \, \text{J} = 6.242 \times 10^{12} \, \text{MeV} \). Thus, the energy is: \[ 1.94283 \times 10^{-13} \, \text{J} \times 6.242 \times 10^{12} \, \text{MeV/J} = 1.213 \, \text{MeV} \]. This is the energy released in MeV.

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Key Concepts

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

Mass Defect
In nuclear physics, the concept of mass defect is crucial to understanding energy release during nuclear decay. Mass defect (\( \Delta m \)) refers to the small difference in mass between the original nucleus and the sum of the masses of its decay products. When cesium-137 (\( {}^{137} \mathrm{Cs} \)) decays into barium-137 (\( {}^{137} \mathrm{Ba} \)), there is a tiny mass difference that accounts for the energy released in the process. In this exercise, the mass defect calculated is\[\Delta m = m_{\text{Cs}} - m_{\text{Ba}} = 0.0013 \ \text{u},\]where atomic mass units (u) provide a useful scale for the subatomic world.
Energy Conversion
Mass-energy equivalence, famously summarized by Einstein's equation \( E = mc^2 \), allows us to understand how a seemingly negligible mass defect can lead to significant energy release. Converting the mass defect from atomic mass units (u) to kilograms is the first step. We use:
  • \( 1 \ \text{u} \approx 1.660539 \times 10^{-27} \ \text{kg} \).
For cesium-137 decay:\[\Delta m = 0.0013 \ \text{u} \times 1.660539 \times 10^{-27} \ \text{kg/u} \approx 2.1587 \times 10^{-30} \ \text{kg}.\]Then, the energy released is computed by multiplying the mass defect by the speed of light squared\[ E = \Delta m \times (3 \times 10^8 \, \text{m/s})^2 = 1.94283 \times 10^{-13} \, \text{J}, \]where Joules (J) serve as the basic unit of energy.
Half-Life
The concept of half-life is central when discussing nuclear decay. It is defined as the time required for half of a sample of radioactive atoms to decay. For cesium-137, the half-life is notably lengthy at 30.2 years. This means that if you start with a certain amount of cesium-137, half of it would decay into barium-137 over 30.2 years. The significance of this for environmental and health concerns is substantial:
  • Long half-life implies prolonged environmental presence.
  • Radioactive contaminants remain active sources of radiation for extended periods.
Understanding half-life helps in predicting how quickly a radioactive substance will diminish in activity and potential impact.
Cesium-137 Decay
Cesium-137 is a well-known fission product of nuclear reactions and bomb detonations, and its decay is an important study in nuclear chemistry. Decay of cesium-137 is primarily beta decay, leading to the formation of barium-137. This transition is represented as:
  • \( {}^{137}\text{Cs} \rightarrow {}^{137}\text{Ba} + \beta^- \).
The decay process releases energy as seen from the conversion of mass defect into energy. Particularly worrying is the fact that cesium-137 is a gamma-emitter, which is highly penetrating and can pose serious health risks upon exposure.This emphasizes the importance of understanding and managing its presence in the environment, especially given its origin from nuclear activities.

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Most popular questions from this chapter

A typical chest \(x\) -ray radiation dose is \(250 \mu \mathrm{Sv},\) delivered by x rays with an RBE factor of \(0.85 .\) Assuming that the mass of the exposed tissue is one-half the patient's mass of \(88 \mathrm{~kg}\), calculate the energy absorbed in joules.

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.)

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) \({ }^{235} \mathrm{U} \rightarrow{ }^{232} \mathrm{Th}+{ }^{3} \mathrm{He},\) (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 $$ \begin{array}{llll} { }^{232} \mathrm{Th} & 232.0381 \mathrm{u} & { }^{3} \mathrm{He} & 3.0160 \mathrm{u} \\ { }^{231} \mathrm{Th} & 231.0363 \mathrm{u} & { }^{4} \mathrm{He} & 4.0026 \mathrm{u} \\ { }^{230} \mathrm{Th} & 230.0331 \mathrm{u} & { }^{5} \mathrm{He} & 5.0122 \mathrm{u} \\ { }^{235} \mathrm{U} & 235.0429 \mathrm{u} & & \end{array} $$

The nuclide \({ }^{14} \mathrm{C}\) contains (a) how many protons and (b) how many neutrons?

The radioactive nuclide \({ }^{99}\) Tc can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99} \mathrm{Mo},\) a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h}\). Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc},\) which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de- excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc} ?\) (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excited-state \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?
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