what happens to the decaying proton during positron emission
eleven.iv: Nuclear Disuse
- Page ID
- 86254
Learning Objectives
- Write and residue nuclear equations
- To know the unlike kinds of radioactive disuse.
- To remainder a nuclear reaction.
Unstable nuclei spontaneously emit radiation in the form of particles and energy. This by and large changes the number of protons and/or neutrons in the nucleus, resulting in a more stable nuclide. One type of a nuclear reaction is radioactive decay , a reaction in which a nucleus spontaneously disintegrates into a slightly lighter nucleus, accompanied by the emission of particles, energy, or both. An example is shown beneath, in which the nucleus of a polonium atom radioactively decays into a lead nucleus.
\[\ce{^{235}_{92}U \rightarrow \, _2^4He + \, _{90}^{231}Th} \label{Eq2}\]
Notation that in a balanced nuclear equation, the sum of the diminutive numbers (subscripts) and the sum of the mass numbers (superscripts) must be equal on both sides of the equation. How do nosotros know that a product of the reaction is \(\ce{_{90}^{231}Thursday}\)? We use a modified blazon of the law of conservation of affair, which says that we must have the same number of protons and neutrons on both sides of the chemical equation. If our uranium nucleus loses two protons from the alpha particle, then there are 90 protons remaining, identifying the chemical element as thorium. Moreover, if we lose iv nuclear particles of the original 235, there are 231 remaining. Thus, we utilise subtraction to identify the isotope of the thorium cantlet—in this instance, \(\ce{^{231}_{90}Th}\).
Considering the number of protons changes as a upshot of this nuclear reaction, the identity of the chemical element changes. Transmutation is a modify in the identity of a nucleus as a upshot of a modify in the number of protons. There are several unlike types of naturally occurring radioactive decay, and we will examine each separately.
Alpha Disuse
An alpha particle \(\left( \alpha \right)\) is a helium nucleus with two protons and ii neutrons. Alpha particles are emitted during some types of radioactive decay. The net charge of an alpha particle is \(2+\), and its mass is approximately \(4 \: \text{amu}\). The symbol for an alpha particle in a nuclear equation is usually \(\ce{^4_2He}\), though sometimes \(\alpha\) is used. Alpha decay typically occurs for very heavy nuclei in which the nuclei are unstable due to big numbers of nucleons. For nuclei that undergo alpha decay, their stability is increased by the subtraction of two protons and two neutrons. For example, uranium-238 decays into thorium-234 by the emission of an blastoff particle (run across figure beneath).
Instance \(\PageIndex{i}\): Radon-222
Write the nuclear equation that represents the radioactivity of radon-222 past alpha particle emission and place the girl isotope.
Solution
Radon has an atomic number of 86, and then the parent isotope is represented equally \(\ce{^{222}_{86}Rn}\). We represent the alpha particle as \(\ce{^{four}_{2}He}\) and use subtraction (222 − 4 = 218 and 86 − ii = 84) to identify the daughter isotope as an isotope of polonium, \(\mathrm{^{218}_{84}Po}\):
\[\ce{_{86}^{222}Rn\rightarrow \, _2^4He + \, _{84}^{218}Po} \nonumber\]
Exercise \(\PageIndex{one}\): Polonium-209
Write the nuclear equation that represents the radioactive decay of polonium-209 by alpha particle emission and identify the girl isotope.
- Respond
-
\[\ce{_{84}^{209}Po\rightarrow \, _2^4He + \, _{82}^{205}Pb} \nonumber\]
Beta Decay
Nuclei above the band of stability are unstable because their neutron to proton ratio is besides high. To decrease that ratio, a neutron in the nucleus is capable of turning into a proton and an electron. The electron is immediately ejected at a high speed from the nucleus. A beta particle \(\left( \beta \correct)\) is a loftier-speed electron emitted from the nucleus of an atom during some kinds of radioactive decay (come across effigy below). The symbol for a beta particle in an equation is either \(\beta\) or \(\ce{^0_{-one}eastward}\). Carbon-14 undergoes beta decay, transmutating into a nitrogen-14 nucleus.
\[\ce{^{fourteen}_6C} \rightarrow \ce{^{14}_7N} + \ce{^0_{-1}east}\]
Note that beta decay increases the diminutive number past 1, only the mass number remains the aforementioned.
Example \(\PageIndex{ii}\): Boron-12
Write the nuclear equation that represents the radioactivity of boron-12 by beta particle emission and identify the girl isotope. A gamma ray is emitted simultaneously with the beta particle.
Solution
The parent isotope is \(\ce{^{12}_{5}B}\) while one of the products is an electron, \(\ce{^{0}_{-1}e}\). So that the mass and diminutive numbers take the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an diminutive number of 6 is carbon. Thus, the consummate nuclear equation is as follows:
\[\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0e + \gamma} \nonumber\]
The girl isotope is \(\ce{^{12}_6 C}\).
Exercise \(\PageIndex{ii}\): Iodine-131
Write the nuclear equation that represents the radioactivity of iodine-131 past beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle.
- Answer
-
\[\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-i}^0e + \gamma} \nonumber\]
Positron Emission
Nuclei below the band of stability are unstable because their neutron to proton ratio is likewise low. Ane manner to increase that ratio is for a proton in the nucleus to plough into a neutron and another particle chosen a positron. A positron is a particle with the same mass as an electron, only with a positive charge. Similar the beta particle, a positron is immediately ejected from the nucleus upon its formation. The symbol for a positron in an equation is \(\ce{^0_{+1}e}\). For example, potassium-38 emits a positron, becoming argon-38.
\[\ce{^{38}_{nineteen}G} \rightarrow \ce{^{38}_{18}Ar} + \ce{^0_1e}\]
Positron emission decreases the diminutive number past one, but the mass number remains the same.
Electron Capture
An alternating way for a nuclide to increase its neutron to proton ratio is by a phenomenon called electron capture. In electron capture, an electron from an inner orbital is captured by the nucleus of the atom and combined with a proton to form a neutron. For example, silver-106 undergoes electron capture to become palladium-106.
\[\ce{^{106}_{47}Ag} + \ce{^0_{-ane}east} \rightarrow \ce{^{106}_{46}Pd}\]
Note that the overall consequence of electron capture is identical to positron emission. The atomic number decreases past i while the mass number remains the same.
Gamma Ray Emission
Gamma rays \(\left( \gamma \right)\) are very loftier energy electromagnetic waves emitted from a nucleus. Gamma rays are emitted by a nucleus when nuclear particles undergo transitions between nuclear energy levels. This is coordinating to the electromagnetic radiations emitted when excited electrons drop from higher to lower energy levels; the only difference is that nuclear transitions release much more than energetic radiation. Gamma ray emission often accompanies the decay of a nuclide by other means.
\[\ce{^{230}_{90}Th} \rightarrow \ce{^{226}_{88}Ra} + \ce{^4_2He} + \gamma\]
The emission of gamma radiations has no effect on the atomic number or mass number of the products, but information technology reduces their energy.
Summary of Nuclear Radiations
The table below summarizes the main types of nuclear radiation, including charge, mass, symbol, and penetrating power. Penetrating power refers to the relative ability of the radiation to pass through common materials. Radiation with loftier penetrating ability is potentially more unsafe because information technology tin can laissez passer through skin and exercise cellular damage.
| Type | Symbol | Mass number | Charge | Penetration Power | Shielding |
|---|---|---|---|---|---|
| Blastoff particle | \(\ce{^4_2He}\) or \( \alpha \) | 4 | \(2+\) | Depression | Paper, skin |
| Beta particle | \(\ce{^0_{-ane}e}\) or \( \beta \) | 0 | \(1-\) | Moderate | Heavy cloth, plastic |
| Positron | \(\ce{^0_1e}\) or \( \beta^+ \) | 0 | \(one+\) | Moderate | Heavy cloth, plastic |
| Gamma ray | \(\gamma\) or \(^0_0\gamma\) | 0 | 0 | High | Lead, physical |
| Neutron | \(\ce{^1_0n}\) | 1 | 0 | Loftier | Water, lead |
Example \(\PageIndex{iii}\)
Write a counterbalanced nuclear equation to draw each reaction.
- the beta decay of \(^{35}_{16}\textrm{Due south}\)
- the decay of \(^{201}_{80}\textrm{Hg}\) by electron capture
- the disuse of \(^{thirty}_{15}\textrm{P}\) by positron emission
Given: radioactive nuclide and manner of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify whatever missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and ane of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and i of the products and indicates the unknown production as \(^{A}_{Z}\textrm{X}\): \[^{35}_{xvi}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber\]
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The chemical element with Z = 17 is chlorine, so the counterbalanced nuclear equation is equally follows: \[^{35}_{xvi}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber\]
b.
A We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{east}\). The reaction is every bit follows: \[^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber\]
B Both protons and neutrons are conserved, and then the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the production must be Z = 80 + (−1) = 79, which corresponds to the element gilded. The balanced nuclear equation is thus \[^{201}_{lxxx}\textrm{Hg}+\,^{0}_{-1}\textrm eastward\rightarrow\,^{201}_{79}\textrm{Au} \nonumber\]
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this instance, a positron. The unbalanced nuclear equation is therefore \[^{xxx}_{fifteen}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+i}\beta \nonumber\]
B The mass number of the second production is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \[^{30}_{15}\textrm{P}\rightarrow\,^{thirty}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber\]
Practise \(\PageIndex{3}\)
Write a balanced nuclear equation to depict each reaction.
- \(^{11}_{6}\textrm{C}\) past positron emission
- the beta disuse of molybdenum-99
- the emission of an α particle followed past gamma emission from \(^{185}_{74}\textrm{Due west}\)
- Answer a
-
\(^{11}_{6}\textrm{C}\rightarrow\,^{eleven}_{v}\textrm{B}+\,^{0}_{+one}\beta\)
- Respond d
-
\(^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-ane}\beta\)
- Answer c
-
\(^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{ii}\alpha +\,^{0}_{0}\gamma\)
Example \(\PageIndex{4}\)
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
- \(^{45}_{22}\textrm{Ti}\)
- \(^{242}_{94}\textrm{Pu}\)
- \(^{12}_{5}\textrm{B}\)
- \(^{256}_{100}\textrm{Fm}\)
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
- This nuclide has a neutron-to-proton ratio of just 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios disuse past converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this instance, both are observed, with positron emission occurring about 86% of the time and electron capture most fourteen% of the time.
- Nuclei with Z > 83 are too heavy to exist stable and commonly undergo alpha decay, which decreases both the mass number and the diminutive number. Thus \(^{242}_{94}\textrm{Pu}\) is expected to decay by alpha emission.
- This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light chemical element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increment in the atomic number with no change in the mass number. We therefore predict that \(^{12}_{5}\textrm{B}\) will undergo beta decay.
- This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to disuse by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that \(^{256}_{100}\textrm{Fm}\) will disuse by either or both of these two processes. In fact, it decays past both spontaneous fission and alpha emission, in a 97:3 ratio.
Practise \(\PageIndex{4}\)
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
- \(^{32}_{14}\textrm{Si}\)
- \(^{43}_{21}\textrm{Sc}\)
- \(^{231}_{91}\textrm{Pa}\)
- Answer a
-
beta decay
- Answer d
-
positron emission or electron capture
- Answer c
-
alpha disuse
Contributors and Attributions
-
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
Source: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.04%3A_Nuclear_Decay
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