# A Beta Particle May Be Spontaneously Emitted From

A Beta Particle May Be Spontaneously Emitted From

## xi.4: Nuclear Decay

##### Learning Objectives

• Write and balance nuclear equations.

Unstable nuclei spontaneously emit radiation in the course of particles and energy. This by and large changes the number of protons and/or neutrons in the nucleus, resulting in a more than stable nuclide. One type of a nuclear reaction is
, 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 below, in which the nucleus of a polonium atom radioactively decays into a lead nucleus.

$\ce{^{235}_{92}U \rightarrow \, _2^4He + \, _{90}^{231}Th} \label{Eq2}$

Annotation that in a balanced nuclear equation, the sum of the atomic numbers (subscripts) and the sum of the mass numbers (superscripts) must be equal on both sides of the equation. How do we know that a product of the reaction is $$\ce{_{90}^{231}Th}$$? Nosotros utilise a modified type of the
law of conservation of affair, which says that nosotros must have the same number of protons and neutrons on both sides of the chemic equation. If our uranium nucleus loses ii protons from the alpha particle, so there are 90 protons remaining, identifying the element as thorium. Moreover, if we lose four nuclear particles of the original 235, there are 231 remaining. Thus, nosotros use subtraction to identify the isotope of the thorium cantlet—in this case, $$\ce{^{231}_{90}Th}$$.

Because the number of protons changes every bit a result of this nuclear reaction, the identity of the chemical element changes.
Transmutation

is a change in the identity of a nucleus as a result of a change in the number of protons. There are several different types of naturally occurring radioactive disuse, and we will examine each separately.

## Alpha Emission

An
alpha particle $$\left( \alpha \right)$$
is a helium nucleus with two protons and two neutrons.
Blastoff particles are emitted during some types of radioactivity. The net charge of an alpha particle is $$2+$$, and its mass is approximately $$4 \: \text{amu}$$. The symbol for an blastoff particle in a nuclear equation is usually $$\ce{^4_2He}$$, though sometimes $$\alpha$$ is used. Alpha emission 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 past the subtraction of 2 protons and two neutrons. For case, uranium-238 decays into thorium-234 by the emission of an alpha particle (run across Figure $$\PageIndex{one}$$). Figure $$\PageIndex{1}$$: The unstable uranium-238 nucleus spontaneously decays into a thorium-234 nucleus past emitting an alpha particle.
##### Instance $$\PageIndex{1}$$: Radon-222

Write the nuclear equation that represents the radioactive decay of radon-222 by alpha particle emission and place the daughter isotope.

##### Do $$\PageIndex{ane}$$: Polonium-209

Write the nuclear equation that represents the radioactive decay of polonium-209 past alpha particle emission and identify the daughter isotope.

$\ce{_{84}^{209}Po\rightarrow \, _2^4He + \, _{82}^{205}Lead} \nonumber$

## Beta Emission

Nuclei above the band of stability are unstable considering their neutron to proton ratio is too loftier. To subtract 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 $$\left( \beta \right)$$
is a high-speed electron emitted from the nucleus of an atom during some kinds of radioactive decay
(see Figure $$\PageIndex{2}$$). The symbol for a beta particle in an equation is either $$\beta$$ or $$\ce{^0_{-1}due east}$$. Carbon-14 undergoes beta decay, transmutating into a nitrogen-14 nucleus.

$\ce{^{14}_6C} \rightarrow \ce{^{14}_7N} + \ce{^0_{-1}east}$

Annotation that beta decay increases the diminutive number by one, but the mass number remains the same. Figure $$\PageIndex{2}$$: The beta emission of a carbon-fourteen nuclide involves the conversion of a neutron to a proton and an electron, with the electron being emitted from the nucleus.
##### Example $$\PageIndex{2}$$: Boron-12

Write the nuclear equation that represents the radioactive disuse of boron-12 past beta particle emission and place the daughter isotope. A gamma ray is emitted simultaneously with the beta particle.

###### Solution

The parent isotope is $$\ce{^{12}_{five}B}$$ while one of the products is an electron, $$\ce{^{0}_{-one}e}$$. 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 atomic 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 $$\ce{^{12}_6 C}$$.

##### Practise $$\PageIndex{2}$$: Iodine-131

Write the nuclear equation that represents the radioactive decay of iodine-131 by beta particle emission and identify the girl isotope. A gamma ray is emitted simultaneously with the beta particle.

$\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-one}^0e + \gamma} \nonumber$

## Gamma Emission

Gamma rays $$\left( \gamma \right)$$

are very high energy electromagnetic waves emitted from a nucleus.
Gamma rays are emitted past a nucleus when nuclear particles undergo transitions betwixt nuclear energy levels. This is analogous to the electromagnetic radiations emitted when excited electrons drop from higher to lower free energy levels; the only difference is that nuclear transitions release much more than energetic radiation. Gamma ray emission oftentimes accompanies the disuse of a nuclide by other means.

$\ce{^{230}_{90}Th} \rightarrow \ce{^{226}_{88}Ra} + \ce{^4_2He} + \gamma$

The emission of gamma radiation has no outcome on the atomic number or mass number of the products, merely it reduces their energy.

## Positron Emission

Nuclei below the band of stability are unstable because their neutron to proton ratio is too low. One way to increase that ratio is for a proton in the nucleus to plough into a neutron and another particle called a positron. A
positron
is a particle with the same mass equally an electron, just with a positive accuse. 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.

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$\ce{^{38}_{nineteen}Yard} \rightarrow \ce{^{38}_{xviii}Ar} + \ce{^0_1e}$

Positron emission decreases the atomic number by 1, but the mass number remains the same.

## Electron Capture

An alternate way for a nuclide to increase its neutron to proton ratio is by a phenomenon called
electron capture, sympolized Due east.C. In electron capture, an electron from an inner orbital is captured by the nucleus of the cantlet and combined with a proton to course a neutron. For case, silver-106 undergoes electron capture to become palladium-106.

$\ce{^{106}_{47}Ag} + \ce{^0_{-i}east} \rightarrow \ce{^{106}_{46}Pd}$

Note that the overall result of electron capture is identical to positron emission. The atomic number decreases by one while the mass number remains the same.

Tabular array $$\PageIndex{i}$$ lists the characteristics of the different types of radioactive decay.

Table $$\PageIndex{one}$$
Type Symbol Alter in Atomic Number Change in Mass Number Modify in Number of Neutrons
Blastoff emission $$\ce{^4_2He}$$ or $$\blastoff$$ –two –4 –ii
Beta emission $$\ce{^0_{-ane}e}$$ or $$\beta$$ +1 –ane
Gamma emission $$\gamma$$ or $$^0_0\gamma$$
Positron emission $$\ce{^0_1e}$$ or $$\beta^+$$ –1 +1
electron capture Due east.C. –i +one
##### Instance $$\PageIndex{3}$$

Write a balanced nuclear equation to describe each reaction.

1. the beta decay of $$^{35}_{sixteen}\textrm{S}$$
2. the disuse of $$^{201}_{fourscore}\textrm{Hg}$$ past electron capture
3. the disuse of $$^{30}_{fifteen}\textrm{P}$$ past positron emission

Given:
radioactive nuclide and mode of decay

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 residuum the equation.

###### Solution

a.

A
We know the identities of the reactant and one of the products (a β particle). We can therefore brainstorm past writing an equation that shows the reactant and one of the products and indicates the unknown product as $$^{A}_{Z}\textrm{X}$$: $^{35}_{xvi}\textrm{Southward}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-i}\beta \nonumber$

B
Considering 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 − (−ane) = 17. The element with
Z
= 17 is chlorine, and then the counterbalanced nuclear equation is as follows: $^{35}_{sixteen}\textrm{South}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$

b.

A
We know the identities of both reactants: $$^{201}_{80}\textrm{Hg}$$ and an inner electron, $$^{0}_{-1}\textrm{e}$$. The reaction is as follows: $^{201}_{fourscore}\textrm{Hg}+\,^{0}_{-1}\textrm eastward\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$

B
Both protons and neutrons are conserved, so the mass number of the product must be
A
= 201 + 0 = 201, and the atomic number of the product must exist
Z
= lxxx + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{lxxx}\textrm{Hg}+\,^{0}_{-1}\textrm eastward\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$

c.

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A
As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+i}\beta \nonumber$

B
The mass number of the second product is
A
= 30 − 0 = 30, and its atomic number is
Z
= xv − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{xv}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+ane}\beta \nonumber$

##### Exercise $$\PageIndex{3}$$

Write a counterbalanced nuclear equation to describe each reaction.

1. $$^{11}_{6}\textrm{C}$$ by positron emission
2. the beta disuse of molybdenum-99
3. the emission of an α particle followed by gamma emission from $$^{185}_{74}\textrm{W}$$

$$^{11}_{half-dozen}\textrm{C}\rightarrow\,^{xi}_{v}\textrm{B}+\,^{0}_{+1}\beta$$

$$^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$$

Respond c

$$^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{ii}\alpha +\,^{0}_{0}\gamma$$

##### Example $$\PageIndex{4}$$

Predict the kind of nuclear modify each unstable nuclide undergoes when it decays.

1. $$^{45}_{22}\textrm{Ti}$$
2. $$^{242}_{94}\textrm{Pu}$$
3. $$^{12}_{5}\textrm{B}$$
4. $$^{256}_{100}\textrm{Fm}$$

Given:
nuclide

type of nuclear disuse

Strategy:

Based on the neutron-to-proton ratio and the value of
Z, predict the type of nuclear decay reaction that volition produce a more stable nuclide.

###### Solution
1. This nuclide has a neutron-to-proton ratio of but 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that take low neutron-to-proton ratios decay by 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 example, both are observed, with positron emission occurring almost 86% of the time and electron capture about xiv% of the time.
2. Nuclei with
Z
> 83 are as well heavy to be stable and usually 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.
3. This nuclide has a neutron-to-proton ratio of 1.iv, which is very high for a lite 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 increase in the diminutive number with no change in the mass number. Nosotros therefore predict that $$^{12}_{v}\textrm{B}$$ will undergo beta decay.
4. 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 decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $$^{256}_{100}\textrm{Fm}$$ will decay past either or both of these two processes. In fact, it decays past both spontaneous fission and alpha emission, in a 97:3 ratio.
##### Exercise $$\PageIndex{iv}$$

Predict the kind of nuclear modify each unstable nuclide undergoes when it decays.

1. $$^{32}_{xiv}\textrm{Si}$$
2. $$^{43}_{21}\textrm{Sc}$$
3. $$^{231}_{91}\textrm{Pa}$$
Respond a

beta decay

positron emission or electron capture