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Chapter 31: Problem 57

Complete the following decay processes by stating what the symbol \(X\) represents \(\left(X=a, \beta^{-}, \beta^{+},\right.\) or \(\left.\gamma\right)\) (a) \({ }^{211} \mathrm{Pb} \rightarrow{ }_{83}^{211} \mathrm{Bi}+\mathrm{X}\) (b) \({ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{B}+\mathrm{X}\) (c) \(\frac{231}{90} T h^{*} \rightarrow \frac{231}{90} T h+X\) (d) \({ }_{84}^{210} \mathrm{Po} \rightarrow \frac{206}{\mathrm{8} 2 \mathrm{}}+\mathrm{X}\)

Short Answer

Expert verified
(a) \(\beta^-\), (b) \(\beta^+\), (c) \(\gamma\), (d) \(a\).

Step by step solution

01

Understanding the Problem

The exercise asks us to identify the type of decay process (using the symbols \(a, \beta^-, \beta^+, \gamma\) for alpha, beta-minus, beta-plus, and gamma decay respectively) in the given nuclear reactions.
02

Analyzing Decay Processes - Problem (a)

For the decay process \(^{211} \mathrm{Pb} \rightarrow_{83}^{211} \mathrm{Bi}+X\), observe the changes in atomic number and mass number. Here, the atomic number increases by 1 (from 82 to 83) while the mass number remains constant at 211. This indicates a beta-minus decay (\(\beta^-\)), which increases the atomic number by converting a neutron into a proton.
03

Analyzing Decay Processes - Problem (b)

In the decay process \(^{11}_{6} \mathrm{C} \rightarrow ^{11}_{5} \mathrm{B}+X\), the atomic number decreases by 1 (from 6 to 5), but mass number remains the same. This suggests a beta-plus decay (\(\beta^+\)), which decreases the atomic number by converting a proton into a neutron.
04

Analyzing Decay Processes - Problem (c)

For \(\frac{231}{90} T h^{*} \rightarrow \frac{231}{90} T h+X\), both the atomic number and mass number remain unchanged. The symbol \(^{*}\) indicates an excited state of \(Th\), and the decay results in the same element \(Th\) without changes in atomic or mass numbers. This is a gamma decay (\(\gamma\)), where energy is emitted via a gamma photon.
05

Analyzing Decay Processes - Problem (d)

In the reaction \({ }_{84}^{210} \mathrm{Po} \rightarrow \frac{206}{82}+X\), both the atomic number decreases by 2 (from 84 to 82) and the mass number decreases by 4 (from 210 to 206). This indicates an alpha decay (\(a\)), characterized by the emission of an alpha particle \(\left(^{4}_{2}\mathrm{He}\right)\).

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Key Concepts

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

Alpha Decay
Alpha decay is a type of nuclear decay where an unstable nucleus emits an alpha particle, consisting of two protons and two neutrons. This particle is essentially a helium nucleus, denoted as \(^{4}_{2} ext{He}\). As a result, the parent nucleus loses 2 protons and 4 units of mass number.
This process reduces the atomic number by 2 and the mass number by 4, creating a different element.
Alpha decay occurs mainly in heavy elements, such as uranium and radium.
  • Decreases atomic number by 2.
  • Decreases mass number by 4.
  • Common in elements with high atomic numbers.
Beta-Minus Decay
In beta-minus decay, a neutron inside the nucleus transforms into a proton while emitting an electron (known as a beta particle) and an antineutrino. This process increases the atomic number by 1, since a new proton is formed, but the mass number remains unchanged.
Beta-minus decay often occurs in neutron-rich nuclei where the balance of neutrons and protons alters.
The emitted electron is referred to as a \(\beta^-\) particle.
  • Increases atomic number by 1.
  • Mass number stays the same.
  • Occurs in neutron-rich nuclei.
Beta-Plus Decay
Beta-plus decay involves the conversion of a proton into a neutron with the emission of a positron and a neutrino. This decreases the atomic number by 1 without altering the mass number.
This form of decay is typical in proton-rich nuclei where there are more protons than neutrons. The emitted positron is the \(\beta^+\) particle.
  • Decreases atomic number by 1.
  • Mass number remains constant.
  • Common in proton-rich nuclei.
Gamma Decay
Gamma decay occurs when an excited nucleus releases excess energy in the form of gamma rays, which are high-energy photons. This decay does not involve any change in atomic or mass numbers, as no particles are emitted.
Gamma rays accompanying other types of decay serve to stabilize the nucleus without transforming it to a different element.
  • No change in atomic number or mass number.
  • Energy emitted as gamma radiation.
  • Stabilizes the nucleus.

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Most popular questions from this chapter

A Radioisotope Thermoelectric Generator. You and your team are designing a thermoelectric generator that converts the thermal energy released during the decay of a radioactive material into electrical energy. Such devices are used in deep-space probes such as Voyager \(1,\) which was launched in \(1977 .\) Its radioisotope thermoelectric generators are expected to function until the year \(2025 .\) The isotope that your team will use is \({ }^{2}{ }_{94}^{28} \mathrm{Pu}\) (atomic mass \(=238.049553\) u), which has a half-life of 87.7 years. (a) If 'gi Pu undergoes alpha decay, what is the daughter nucleus? (b) Assuming the daughter nucleus found in (a) has an atomic mass of \(234.0409468 \mathrm{u}\) calculate the energy released in the decay in joules (the atomic weight of an alpha particle is 4.002603 u). (c) How many atoms are in one gram of \(\frac{238}{94}\) Pu? (d) How many atoms in one gram of \({ }^{238}\) Pu decay after one year? (e) How much energy is released from one gram of \({ }_{94}^{238}\) Pu during one year? (f) What is the average power output of one gram during one year (in watts)?

$$ \text { In the form } \frac{A}{Z} X, \text { identify the daughter nucleus that results when } $$ (a) plutonium \({ }_{94}^{242}\) Pu undergoes \(\alpha\) decay, (b) sodium \(_{11}^{24}\) Na undergoes \(\beta^{-}\) decay, and (c) nitrogen \({ }_{7}^{13} \mathrm{N}\) undergoes \(\beta^{+}\) decay.

(a) Energy is required to separate a nucleus into its constituent nucleons, as Interactive Figure 31.3 indicates; this energy is the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single nucleon to the remainder of the nucleus. For example, separating nitrogen \({ }^{14}{ }_{7} \mathrm{N}\) into nitrogen \({ }^{13}{ }_{7} \mathrm{N}\) and a neutron takes energy equal to the binding energy of the neutron, as shown below: Find the energy (in MeV) that binds the neutron to the \({ }^{14}, \mathrm{N}\) nucleus by considering the mass of \({ }_{7}^{13} \mathrm{N}\) (atomic mass \(=13.005738 \mathrm{u}\) ) and the mass of \({ }_{0}^{1}\) n (atomic mass \(=1.008665 \mathrm{u}\) ), as compared to the mass of \({ }_{7}^{14} \mathrm{N}\) (atomic mass \(=\) \(14.003074 \mathrm{u}) .\) (b) Similarly, one can speak of the energy that binds a single proton to the \({ }^{14}{ }_{7} \mathrm{N}\) nucleus: Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton (atomic mass \(=1.007825 \mathrm{u}\) ) to the \({ }_{7}^{14} \mathrm{N}\) nucleus. The atomic mass of carbon \({ }_{6}^{13} \mathrm{C}\) is \(13.003355 \mathrm{u}\). (c) Which nucleon is more tightly bound, the neutron or the proton?

A device used in radiation therapy for cancer contains \(0.50 \mathrm{g}\) of cobalt \({ }_{27}^{60} \mathrm{Co}(59.933819 \mathrm{u}) .\) The half-life of \({ }_{27}^{60} \mathrm{Co}\) is \(5.27 \mathrm{yr} .\) Determine the activity of the radioactive material.

Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are three times as many A nuclei as there are \(\mathrm{B}\) nuclei. The half-life of species \(\mathrm{B}\) is 1.50 days. Find the half-life of species A.
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