Gamma ray emission accompanies either alpha decay or beta decay.

A nuclear decay equation shows the nucleus of the unstable isotope (parent) on the left hand side of the equation and the type of radiation emitted as well as the new nucleus (daughter) is shown on the right hand side of the equation:

⚛ element X undergoes alpha decay to produce element Y

parent nucleus

→

radiation

+

daughter nucleus

A

X

Z

→

4

He

2

+

(A-4)

Y

(Z-2)

⚛ element X undergoes beta decay to produce element Z

parent nucleus

→

radiation

+

daughter nucleus

A

X

Z

→

0

e

-1

+

A

Z

(Z+1)

Mass numbers (A) on the left hand side of the nuclear decay equation must equal the sum of the mass numbers on the right hand side of the equation.

Atomic numbers (Z) on the left hand side of the nuclear decay equation must equal the sum of the atomic numbers on the right hand side of the equation.

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Alpha Decay

Imagine that, on the day you are born, Chris the Chemist gives you a special present, 100.0 grams of radium-226 sealed in a box with a nitrogen gas atmosphere.
On your 16^{th} birthday you open the box and when you weigh the lump of radium-226 you find there is only 99.3 grams left.
You refill the box with nitrogen gas before sealing it closed.
On your 32^{nd} birthday, you open the box, catch the gas that escapes, have it tested and find that while nitrogen is still present there is also radon-222, then weigh the remaining radium-226 only to find that it now only weighs 98.6 grams.

Is someone stealing your radium or has some radium-226 turned into radon-222?

Consider the nucleus of each of these atoms.
The nucleus of an atom of radium-226 contains 226 nucleons (neutrons and protons).
The nucleus of an atom of radon-222 contains 222 nucleons (neutrons and protons).
We also know that the atomic number (Z) of radium-226 from the periodic table is 88 which means that an atom of radium-226 has 88 protons in its nucleus.
Radon-222 has an atomic number (Z) of 86 so an atom of radon-222 contains 86 protons in its nucleus.

We could write a partial nuclear decay equation to describe what we are seeing:

radium-226

→

radon-222

226

Ra

88

→

222

Rn

86

The nucleus of the radium-226 atom has lost 4 nucleons to become radon-222:

A(radium) - A(radon) = 226 - 222 = 4

Two of the nucleons lost were protons:

Z(radium) - Z(radon) = 88 - 86 = 2

We now have a description of the particle lost as a nucleus that contains 4 nucleons, two of the nucleons are protons.
If you look in the periodic table, you will find that helium has atomic number (Z) 2, that is, the nucleus of a helium atom contains 2 protons.
Therefore we can identify the lost particle as a helium nucleus (or alpha particle), and give it the chemical symbol He.

Now we can complete our equation for the nuclear decay of radium-226 to produce radon-222:

parent nucleus

→

radiation

+

daughter nucleus

radium-226

→

alpha particle

+

radon-222

226

Ra

88

→

4

He

2

+

222

Rn

86

Note that if we add together the mass numbers on the left hand side of the equation we get the same value as for the mass number on the right hand side of the equation:

and that if we add together the atomic numbers on the left hand side of the equation we get the same value as for the atomic number on the right hand side of the equation:

The number of protons and the number of neutrons has been conserved during alpha decay.

Only isotopes of elements with atomic numbers greater than 82 ( Z > 82) undergo alpha decay.
Some examples of these isotopes are given in the table below:

Imagine that you give Chris the Chemist 100.0 grams of iodine-131 as a birthday present, sealed inside a box with a nitrogen atmosphere.
When Chris weighs the iodine sample 8 days later, there is only 50 grams of iodine-131 left, but, when the gas from the box is tested it contains not only nitrogen gas but also xenon-131 gas!

Half the mass of iodine-131 has turned into xenon-131.

Consider the nucleus of each of these atoms:

Element

Mass Number (number of nucleons)

Atomic Number (number of protons)

iodine-131 ^{131}I

131

53

xenon-131 ^{131}Xe

131

54

The number of nucleons, that is the number neutrons + the number of protons, has been conserved, BUT, the nucleus of an atom of xenon has one more proton than the nucleus of an atom of iodine.

One of the neutrons in the nucleus of the iodine-131 atom has changed into a proton.
How could that happen?

A neutron (n) has a mass number of 1 (A = 1), but it has no charge, no protons, so its atomic number is 0 (Z = 0)
A proton (p) has a mass number of 1 (A = 1), but it is a proton so its atomic number is 1 (Z = 1)
We can write a partial equation to describe the conversion of a neutron to a proton as:

neutron

→

proton

1

n

0

→

1

p

1

We can determine the mass number and atomic number of our missing particle:

Mass numbers on left hand side of equation = mass numbers on right hand side of equation:

1 = 1 + A

A = 0

missing particle has mass number (A) = 0

Atomic numbers on left hand side of equation = atomic numbers on right hand side of equation:

0 = 1 + Z

Z = -1

missing particle has atomic number (Z) = -1

So the missing particle has no nucleons (A = 0), neither neutrons nor protons, BUT, it has an atomic number with the same magnitude but the opposite sign to a proton.
This is the description of an electron (e), that is, a particle that is not a nucleon but has negative charge rather than a positive charge like a proton.
Now we can complete our equation to describe how a neutron can change into a proton and an electron (which is called a beta particle because it is produced in the nucleus):^{2}

neutron

→

proton

+

beta particle

1

n

0

→

1

p

1

+

0

e

-1

We can write an equation to describe the nuclear decay of iodine-131 to produce a beta particle and xenon-131:

parent nucleus

→

daughter nucleus

+

radiation

iodine-131

→

xenon-131

+

beta particle

131

I

53

→

131

Xe

54

+

0

e

-1

The total of the mass numbers on the right hand side of the equation = mass numbers on left hand side of equation:

A(Xe) + A(beta particle) = A(I)

131+ 0 = 131

The total of the atomic numbers on the right hand side of the equation = atomic numbers on left hand side of equation:

Z(Xe) + Z(beta particle) = Z(I)

54+ -1 = 53

The total number of nucleons has been conserved during beta decay.

Unstable isotopes of many elements undergo beta decay.
Some examples of these isotopes can be found in the table below:

Worked Examples of Writing Nuclear Decay Equations

Question 1:
Uranium-238 is an unstable isotope of uranium.

What element is produced when uranium-238 undergoes alpha decay?

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?

Identify the element resulting from alpha decay of uranium-238.

What data (information) have you been given in the question?

Extract the data from the question:

Parent isotope: uranium-238

symbol: ^{238}U

Type of radiation emitted: alpha particle

identity of alpha particle:

4

He

2

What is the relationship between what you know and what you need to find out?

(i) Use the periodic table to find the atomic number of uranium (U):

Z = 92

(ii) Write the symbol representing the nucleus of the uranium-238 atom:

238

U

92

(iii) Let X be the chemical symbol representing the unknown daughter isotope.
Write an equation to represent the alpha decay of uranium-238

parent nucleus

→

radiation

+

daughter nucleus

uranium-238

→

alpha particle

+

element X

238

U

92

→

4

He

2

+

(238-4 = 234)

X

(92-2 = 90)

Identify the daughter isotope:

The daughter isotope (element X) has atomic number (Z) 90.

From the periodic table, the element with atomic number 90 is thorium, symbol Th.

Is your answer plausible?

(i) Check that the mass numbers on the left hand side of the equation are equal to the sum of the mass numbers on the right hand side of the equation:

238 = 234 + 4

(ii) Check that the atomic numbers on the left hand side of the equation are equal to the sum of the atomic numbers on the right hand side of the equation:

92 = 90 + 2

Since the number of nucleons has been conserved, we are reasonably confident that our unknown element has atomic number 90.
Check that the atomic number of thorium (Th) is 90 in the periodic table.

State your solution to the problem "element produced by alpha decay of uranium-238":

Element is thorium.

Symbol of this isotope is:

234

Th

90

Question 2:
Carbon-14 is used to date archeological artefacts because it is an unstable isotope of carbon that undergoes nuclear decay by emitting a beta particle.

Write an equation to represent the beta decay of carbon-14.

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?

Write an equation for the beta decay of carbon-14.

What data (information) have you been given in the question?

Extract the data from the question:

Parent isotope: carbon-14

symbol ^{14}C

Type of radiation: beta particle

identity of beta particle:

0

e

-1

What is the relationship between what you know and what you need to find out?

(i) Use the periodic table to find the atomic number of carbon (C):

Z = 6

(ii) Write the symbol representing the nucleus of the carbon-14 atom:

14

C

6

(iii) Let X be the chemical symbol representing the unknown daughter isotope.
Write an equation to represent the beta decay of carbon-14

parent nucleus

→

radiation

+

daughter nucleus

carbon-14

→

beta particle

+

element X

14

C

6

→

0

e

-1

+

(14 - 0 = 14)

X

(6 - -1 = 7)

Identify the daughter isotope and write the equation:

The unknown element has atomic number (Z) 7.

From the periodic table, nitrogen (N) has atomic number 7.

Replace X in the equation with N and calculate its mass number:

parent nucleus

→

radiation

+

daughter nucleus

carbon-14

→

beta particle

+

element X

14

C

6

→

0

e

-1

+

(14 - 0 = 14)

X

(6 - -1 = 7)

14

C

6

→

0

e

-1

+

14

N

7

Is your answer plausible?

(i) Check that the mass numbers on the left hand side of the equation are equal to the sum of the mass numbers on the right hand side of the equation:

14 = 0 + 14

(ii) Check that the atomic numbers on the left hand side of the equation are equal to the sum of the atomic numbers on the right hand side of the equation:

6 = -1 + 7

Since the number of nucleons has been conserved, we are reasonably confident that our unknown element has atomic number 7 and mass number 14.
Check that the atomic number of nitrogen (N) is 7 in the periodic table.

State your solution to the problem "equation for the beta decay of carbon-14":

1. Beta particles, high energy electrons, are emitted when a neutron decays to form a proton and an electron.
There are 2 types of beta decay, β^{-} and β^{+}, where β^{-} represents an electron and β^{+} represents a positron.
β^{-} decay is accompanied by the emission of an antineutrino, β^{+} decay is accompanied by the emission of a neutrino.

2. A neutron consists of 2 down quarks (d) and 1 up quark (u).
A proton consists of 1 down quark (d) and 2 up quarks (u).
The process of beta decay is the process of turning one down quark (d) into an up quark (u):
d → u + β^{-} + antineutrino