Chemistry

7.5 - Nuclear Equations

6 min

note 1 of 5

Writing and balancing a nuclear equation

A nuclear equation shows what a nucleus turns into and what it emits during radioactive decay.
It's written like a chemical equation: the nucleus before decay on the left, an arrow, then the new nucleus plus the emitted particle or radiation.
If a nucleus absorbs a particle, that particle goes on the left.
Every species is written in nuclide notation: mass number at the top left, atomic number at the bottom left (e.g. $_{6}^{14} C$ ).

Emitted particles

Alpha particle - $_{2}^{4} He$ (mass number 4, atomic number 2)
Beta particle - $_{- 1}^{0} e$ (mass number 0, atomic number -1)
Proton - $_{1}^{1} p$ (mass number 1, atomic number 1)
Neutron - $_{0}^{1} n$ (mass number 1, atomic number 0)

Balancing the equation

The total mass number (top) on the left must equal the total mass number on the right.
The total atomic number (bottom) on the left must equal the total atomic number on the right.

Finding an unknown species

Work out the missing mass number by balancing the top row.
Work out the missing atomic number by balancing the bottom row.
If it's a particle, match the numbers to the list above. If it's a nucleus, use the atomic number to look up the element on the periodic table.

note 2 of 5

What each emission does to the nucleus

Emission	Written as	Mass number	Atomic number	What is left behind
Alpha particle	$_{2}^{4} He$	falls by 4	falls by 2	a different element, two places back in the periodic table
Beta particle	$_{- 1}^{0} e$	unchanged	rises by 1	a different element, one place further along
Neutron	$_{0}^{1} n$	falls by 1	unchanged	a lighter isotope of the same element
Proton	$_{1}^{1} p$	falls by 1	falls by 1	a different element, one place back
Gamma ray	γ	unchanged	unchanged	the same nucleus with less energy

Alpha emission

An alpha particle is two protons and two neutrons (a helium nucleus).
The mass number falls by 4 and the atomic number falls by 2, forming a different element.

A nucleus of blue neutrons and green protons labelled original element, an alpha particle of two protons and two neutrons labelled helium nucleus, He with 4 above 2, plus a nucleus labelled new element.

Beta emission

A neutron inside the nucleus turns into a proton and an electron; the electron is ejected as a beta particle.
The mass number is unchanged and the atomic number rises by 1, forming a different element:

$_{6}^{14} C$ → $_{7}^{14} N$ + $_{- 1}^{0} e$

A carbon-14 nucleus with a zoomed-in box showing a neutron turning into a proton and an electron, then an arrow to a beta particle, written e with 0 above -1, and a nitrogen-14 nucleus.

Neutron and proton emission

Neutron emission - The mass number falls by 1 and the atomic number is unchanged, leaving a lighter isotope of the same element.
Proton emission - The mass number falls by 1 and the atomic number falls by 1, forming a different element.

Two panels: neutron emission, where beryllium-13 gives a neutron plus beryllium-12 labelled lighter isotope, and proton emission, where aluminium-20 with atomic number 13 gives a proton plus magnesium-19 labelled different element.

Gamma emission

Gamma radiation is high-energy electromagnetic radiation, not a particle.
Emitting a gamma ray releases excess energy without changing the mass number or atomic number:

$_{90}^{234} Th$ → $_{90}^{234} Th$ + γ

note 3 of 5

Worked example - Writing a decay equation

Uranium-238 decays by alpha emission. Write the balanced nuclear equation for the decay.

Step 1 - Write the decaying nucleus

Uranium has atomic number 92, and uranium-238 has mass number 238:

$_{92}^{238} U$

Step 2 - Take away the alpha particle's numbers

An alpha particle is $_{2}^{4} He$ , so it removes 4 from the mass number and 2 from the atomic number:

mass number: 238 - 4 = 234
atomic number: 92 - 2 = 90

Step 3 - Name the new nucleus and write the equation

The element with atomic number 90 is thorium (Th), so the nucleus left behind is $_{90}^{234} Th$ :

$_{92}^{238} U$ → $_{90}^{234} Th$ + $_{2}^{4} He$

Step 4 - Check both totals

mass numbers: 238 = 234 + 4
atomic numbers: 92 = 90 + 2

Both sides match, so the equation is balanced.

note 4 of 5

Worked example - Identifying an unknown particle

A nitrogen-14 nucleus absorbs a neutron and forms a carbon-14 nucleus and one other particle, X. Identify X.

$_{7}^{14} N$ + $_{0}^{1} n$ → $_{6}^{14} C$ + X

Step 1 - Add up the mass numbers on the left

14 + 1 = 15

Carbon-14 accounts for 14, so X has a mass number of 15 - 14 = 1.

Step 2 - Add up the atomic numbers on the left

7 + 0 = 7

Carbon has atomic number 6, so X has an atomic number of 7 - 6 = 1.

Step 3 - Name the particle with those two numbers

A particle with mass number 1 and atomic number 1 is $_{1}^{1} p$ , so X is a proton.

note 5 of 5

Worked example - Counting the emissions in a decay series

An atom of thorium-227 decays by a series of alpha emissions to form an atom of lead-211. How many alpha particles are released?

Step 1 - Count using the mass numbers

Thorium-227 starts at 227 and lead-211 ends at 211, a fall of 227 - 211 = 16.

Each alpha particle carries away a mass number of 4:

16 ÷ 4 = 4 alpha particles

Step 2 - Count using the atomic numbers

Thorium has atomic number 90 and lead has atomic number 82, a fall of 90 - 82 = 8.

Each alpha particle carries away an atomic number of 2:

8 ÷ 2 = 4 alpha particles

Step 3 - State the answer

Both counts agree, so 4 alpha particles are released.

key terms

TermDefinition

Nuclear equationAn equation that shows what a nucleus turns into and what it gives out during radioactive decay.

end of notes

Key points to remember

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7.6 - Half-Life & Uses of RadioisotopesNext up

Chemistry

7.5 - Nuclear Equations

6 min

note 1 of 5

Writing and balancing a nuclear equation

A nuclear equation shows what a nucleus turns into and what it emits during radioactive decay.
It's written like a chemical equation: the nucleus before decay on the left, an arrow, then the new nucleus plus the emitted particle or radiation.
If a nucleus absorbs a particle, that particle goes on the left.
Every species is written in nuclide notation: mass number at the top left, atomic number at the bottom left (e.g. $_{6}^{14} C$ ).

Emitted particles

Alpha particle - $_{2}^{4} He$ (mass number 4, atomic number 2)
Beta particle - $_{- 1}^{0} e$ (mass number 0, atomic number -1)
Proton - $_{1}^{1} p$ (mass number 1, atomic number 1)
Neutron - $_{0}^{1} n$ (mass number 1, atomic number 0)

Balancing the equation

The total mass number (top) on the left must equal the total mass number on the right.
The total atomic number (bottom) on the left must equal the total atomic number on the right.

Finding an unknown species

Work out the missing mass number by balancing the top row.
Work out the missing atomic number by balancing the bottom row.
If it's a particle, match the numbers to the list above. If it's a nucleus, use the atomic number to look up the element on the periodic table.

note 2 of 5

What each emission does to the nucleus

Emission	Written as	Mass number	Atomic number	What is left behind
Alpha particle	$_{2}^{4} He$	falls by 4	falls by 2	a different element, two places back in the periodic table
Beta particle	$_{- 1}^{0} e$	unchanged	rises by 1	a different element, one place further along
Neutron	$_{0}^{1} n$	falls by 1	unchanged	a lighter isotope of the same element
Proton	$_{1}^{1} p$	falls by 1	falls by 1	a different element, one place back
Gamma ray	γ	unchanged	unchanged	the same nucleus with less energy

Alpha emission

An alpha particle is two protons and two neutrons (a helium nucleus).
The mass number falls by 4 and the atomic number falls by 2, forming a different element.

Beta emission

A neutron inside the nucleus turns into a proton and an electron; the electron is ejected as a beta particle.
The mass number is unchanged and the atomic number rises by 1, forming a different element:

$_{6}^{14} C$ → $_{7}^{14} N$ + $_{- 1}^{0} e$

Neutron and proton emission

Neutron emission - The mass number falls by 1 and the atomic number is unchanged, leaving a lighter isotope of the same element.
Proton emission - The mass number falls by 1 and the atomic number falls by 1, forming a different element.

Gamma emission

Gamma radiation is high-energy electromagnetic radiation, not a particle.
Emitting a gamma ray releases excess energy without changing the mass number or atomic number:

$_{90}^{234} Th$ → $_{90}^{234} Th$ + γ

note 3 of 5

Worked example - Writing a decay equation

Uranium-238 decays by alpha emission. Write the balanced nuclear equation for the decay.

Step 1 - Write the decaying nucleus

Uranium has atomic number 92, and uranium-238 has mass number 238:

$_{92}^{238} U$

Step 2 - Take away the alpha particle's numbers

An alpha particle is $_{2}^{4} He$ , so it removes 4 from the mass number and 2 from the atomic number:

mass number: 238 - 4 = 234
atomic number: 92 - 2 = 90

Step 3 - Name the new nucleus and write the equation

The element with atomic number 90 is thorium (Th), so the nucleus left behind is $_{90}^{234} Th$ :

$_{92}^{238} U$ → $_{90}^{234} Th$ + $_{2}^{4} He$

Step 4 - Check both totals

mass numbers: 238 = 234 + 4
atomic numbers: 92 = 90 + 2

Both sides match, so the equation is balanced.

note 4 of 5

Worked example - Identifying an unknown particle

A nitrogen-14 nucleus absorbs a neutron and forms a carbon-14 nucleus and one other particle, X. Identify X.

$_{7}^{14} N$ + $_{0}^{1} n$ → $_{6}^{14} C$ + X

Step 1 - Add up the mass numbers on the left

14 + 1 = 15

Carbon-14 accounts for 14, so X has a mass number of 15 - 14 = 1.

Step 2 - Add up the atomic numbers on the left

7 + 0 = 7

Carbon has atomic number 6, so X has an atomic number of 7 - 6 = 1.

Step 3 - Name the particle with those two numbers

A particle with mass number 1 and atomic number 1 is $_{1}^{1} p$ , so X is a proton.

note 5 of 5

Worked example - Counting the emissions in a decay series

An atom of thorium-227 decays by a series of alpha emissions to form an atom of lead-211. How many alpha particles are released?

Step 1 - Count using the mass numbers

Thorium-227 starts at 227 and lead-211 ends at 211, a fall of 227 - 211 = 16.

Each alpha particle carries away a mass number of 4:

16 ÷ 4 = 4 alpha particles

Step 2 - Count using the atomic numbers

Thorium has atomic number 90 and lead has atomic number 82, a fall of 90 - 82 = 8.

Each alpha particle carries away an atomic number of 2:

8 ÷ 2 = 4 alpha particles

Step 3 - State the answer

Both counts agree, so 4 alpha particles are released.

key terms

TermDefinition

Nuclear equationAn equation that shows what a nucleus turns into and what it gives out during radioactive decay.

end of notes

Key points to remember

Take a quiz View flashcards

7.6 - Half-Life & Uses of RadioisotopesNext up