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What particle is produced during the following decay processes: (a) sodium-24 decays to magnesium-24; (b) mercury-188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

Short Answer

Expert verified
The particles produced in each decay process are: (a) Sodium-24 decays to Magnesium-24: electron (\(e^-\)) and antineutrino (\(\bar{\nu}_{e}\)). (b) Mercury-188 decays to Gold-188: positron (\(e^+\)) and neutrino (\(\nu_{e}\)) or electron capture. (c) Iodine-122 decays to Xenon-122: electron (\(e^-\)) and antineutrino (\(\bar{\nu}_{e}\)). (d) Plutonium-242 decays to Uranium-238: alpha particle (\(\alpha\)).

Step by step solution

01

Identify the decay processes.

For each decay process, we need to compare the atomic numbers (number of protons) of parent and daughter nuclei. If the atomic number remains the same, it indicates beta decay; if it increases by 1, it indicates beta-minus decay; if it decreases by 1, it indicates beta-plus decay or electron capture; if it decreases by 2, it indicates alpha decay. Step 2 - Determine the particle released in each decay process
02

Determine the particle released in each decay process.

Compare the atomic numbers of the parent and daughter nuclei to find which particle is released: (a) Sodium-24 (Na) has an atomic number of 11, and magnesium-24 (Mg) has an atomic number of 12. Since the atomic number has increased by 1, this decay is \(\beta^-\)decay, and the particle released is an electron (\(e^-\)) and an antineutrino (\(\bar{\nu}_{e}\)). (b) Mercury-188 (Hg) has an atomic number of 80, and gold-188 (Au) has an atomic number of 79. Since the atomic number has decreased by 1, this decay is either \(\beta^+\) decay, releasing a positron (\(e^+\)) and a neutrino (\(\nu_{e}\)), or electron capture, as they have the same net effect on the atomic number. (c) Iodine-122 (I) has an atomic number of 53, and xenon-122 (Xe) has an atomic number of 54. Since the atomic number has increased by 1, this decay is \(\beta^-\)decay, and the particle released is an electron (\(e^-\)) and an antineutrino (\(\bar{\nu}_{e}\)). (d) Plutonium-242 (Pu) has an atomic number of 94, and uranium-238 (U) has an atomic number of 92. Since the atomic number has decreased by 2, this decay is an alpha decay, and the particle released is an alpha particle (\(\alpha\)), which consists of 2 protons and 2 neutrons. Step 3 - Summarize the results
03

Summarize the results.

The particles produced in each decay process are: (a) Sodium-24 decays to Magnesium-24: electron (\(e^-\)) and antineutrino (\(\bar{\nu}_{e}\)). (b) Mercury-188 decays to Gold-188: positron (\(e^+\)) and neutrino (\(\nu_{e}\)) or electron capture. (c) Iodine-122 decays to Xenon-122: electron (\(e^-\)) and antineutrino (\(\bar{\nu}_{e}\)). (d) Plutonium-242 decays to Uranium-238: alpha particle (\(\alpha\)).

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Key Concepts

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

Beta Decay
Beta decay is a type of radioactive decay where unstable nuclei release particles in order to gain stability. There are two types of beta decay: beta-minus decay (\(\beta^-\)) and beta-plus decay (\(\beta^+\)).
  • In beta-minus decay, a neutron is transformed into a proton. This increases the atomic number by one while keeping the mass number constant. Consequently, an electron (\(e^-\)) and an antineutrino (\(\bar{u}_{e}\)) are emitted.
  • In beta-plus decay, a proton is converted into a neutron, which results in a decrease of the atomic number by one. This process emits a positron (\(e^+\)) and a neutrino (\(u_{e}\)).
These fine adjustments in atomic structure allow the nucleus to become more stable. Understanding these transformations helps in grasping how certain decay processes change elements from one to another.
Alpha Decay
Alpha decay is another common type of radioactive decay. In this process, an alpha particle is emitted from the nucleus. An alpha particle consists of 2 protons and 2 neutrons, akin to a helium nucleus. During alpha decay:
  • The atomic number of the element decreases by two because two protons are lost.
  • The mass number decreases by four since two protons and two neutrons are ejected.
This type of decay mostly occurs in heavy elements, like uranium and plutonium, because they benefit from shedding these particles to stabilize their large nuclei. Alpha particles are relatively large and heavy, thus they don't penetrate materials deeply, but they can cause significant ionization in materials they interact with directly.
Atomic Number
The atomic number is a fundamental property of elements, denoted by the letter \(Z\). It represents the number of protons found in the nucleus of an atom.
  • Each element on the periodic table is defined by its unique atomic number.
  • The atomic number determines the element’s chemical properties and its place on the periodic table.
  • In nuclear processes like beta and alpha decay, changes in the atomic number indicate the type of decay that occurs, such as an increase or decrease due to the emission of subatomic particles.
Tracking atomic number changes helps scientists identify the likelihood of an element undergoing radioactive decay and predict the resulting products.
Nuclear Processes
Nuclear processes encompass a variety of phenomena that involve changes in an atom’s nucleus. Both beta and alpha decay are examples of such processes. These processes are vital because:
  • They lead to the transformation of unstable atomic nuclei into more stable configurations.
  • Nuclear processes release energy, which can be harnessed for power in nuclear reactors, or studied for insights into elemental transmutation.
  • They provide valuable information for dating geological formations and archaeological findings through radiometric dating techniques.
    • Understanding nuclear processes involves studying the particles emitted, the resultant changes in nuclei, and the effects on the bulk material. Each process follows specific rules that scientists can predict and calculate.

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Most popular questions from this chapter

Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they decay? Why are alpha emitters not used as diagnostic tools?

A rock contains \(0.257 \mathrm{mg}\) of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is \(4.5 \times 10^{5} \mathrm{yr}\). How old is the rock? SOLUTION Analyze We are told that a rock sample has a certain amount of lead206 for every unit mass of uranium-238 and asked to estimate the age of the rock. Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known half-life. To apply firstorder kinetics expressions (Equations \(21.19\) and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every \(1 \mathrm{mg}\) that remains today. Solve Let's assume that the rock currently contains \(1.000 \mathrm{mg}\) of uranium-238 and therefore \(0.257 \mathrm{mg}\) of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals \(1.000 \mathrm{mg}\) plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add \(1.000 \mathrm{mg}\) and \(0.257 \mathrm{mg}\). We have to multiply the present mass of lead-206 \((0.257 \mathrm{mg})\) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of \({ }_{92}^{239} \mathrm{U}\) was $$ \text { Original } \begin{aligned} { }_{98}^{238} \mathrm{U} &=1.000 \mathrm{mg}+\frac{238}{206}(0.257 \mathrm{mg}) \\ &=1.297 \mathrm{mg} \end{aligned} $$ Using Equation 21.20, we can calculate the decay constant for the process from its half-life: $$ k=\frac{0.693}{4.5 \times 10^{9} \mathrm{yr}}=1.5 \times 10^{-10} \mathrm{yr}^{-1} $$ Rearranging Equation \(21.19\) to solve for time, \(t\), and substituting known quantities gives $$ t=-\frac{1}{k} \ln \frac{N_{t}}{N_{0}}=-\frac{1}{1.5 \times 10^{-10} \mathrm{yr}^{-1}} \ln \frac{1.000}{1.297}=1.7 \times 10^{9} \mathrm{yr} $$ Comment To check this result, you could use the fact that the decay of uranium-235 to lead-207 has a half-life of \(7 \times 10^{8} \mathrm{yr}\) and measure the relative amounts of uranium-235 and lead-207 in the rock.

(a) Which of the following are required characteristics of an isotope to be used as a fuel in a nuclear power reactor? (i) It must emit gamma radiation. (ii) On decay, it must release two or more neutrons. (iii) It must have a half-life less than one hour. (iv) It must undergo fission upon the absorption of a neutron. (b) What is the most common fissionable isotope in a commercial nuclear power reactor?

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{~J}\) of energy and if the average person weighs \(75 \mathrm{~kg}\), calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

Indicate the number of protons and neutrons in the following nuclei: (a) \({ }_{53}^{129} \mathrm{I}\), (b) \({ }^{138} \mathrm{Ba}\), (c) neptunium-237.

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