Conservation of mass and charge

Definition

Balanced decay equation

A decay equation in which the total mass number and the total charge are the same on both sides of the arrow.

A decay equation is balanced when the totals on the two sides of the arrow are the same.
The total mass number on the left must equal the total mass number on the right, because no nucleons are created or destroyed.
The total charge on the left must equal the total charge on the right, because charge is conserved.
The charge of each nucleus is its atomic number, so the bottom numbers must add up to the same total on both sides.
The charge of a beta particle is $- 1$ , so it must be counted as a negative number when the bottom numbers are added.
These two rules let you find any single missing number, nucleus or particle in a decay equation.

Balancing alpha decay

The general equation for alpha decay is $ZAX→Z−2A−4Y+24He{}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\text{He}$ .
The mass numbers balance because $A = (A - 4) + 4$ .
The charges balance because $Z = (Z - 2) + 2$ .
For uranium-235 decaying to thorium-231, the mass numbers balance as $235 = 231 + 4$ and the charges balance as $92 = 90 + 2$ .

Balancing beta decay

The general equation for beta decay is $ZAX→Z+1AY+−10e{}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z+1}\text{Y} + {}^{0}_{-1}\text{e}$ .
The mass numbers balance because $A = A + 0$ .
The charges balance because $Z = (Z + 1) + (- 1)$ .
The daughter nucleus has one more positive charge, and the beta particle carries away one negative charge, so the total charge stays the same.
For thorium-234 decaying to protactinium-234, the mass numbers balance as $234 = 234 + 0$ and the charges balance as $90 = 91 + (- 1)$ .

Balancing gamma and neutron emission

A nucleus with excess energy is often marked with an asterisk, so $ZAX∗{}^{A}_{Z}\text{X}^{*}$ is a nucleus that is about to emit a gamma ray.
The general equation for gamma emission is $ZAX∗→ZAX+00γ{}^{A}_{Z}\text{X}^{*} \rightarrow {}^{A}_{Z}\text{X} + {}^{0}_{0}\gamma$ .
The gamma ray adds $0$ to both totals, so the parent and daughter have the same mass number and the same atomic number.
For example, $4399Tc∗→4399Tc+00γ{}^{99}_{43}\text{Tc}^{*} \rightarrow {}^{99}_{43}\text{Tc} + {}^{0}_{0}\gamma$ shows an excited technetium-99 nucleus losing energy.
The general equation for neutron emission is $ZAX→ZA−1X+01n{}^{A}_{Z}\text{X} \rightarrow {}^{A-1}_{Z}\text{X} + {}^{1}_{0}\text{n}$ , which balances because $A = (A - 1) + 1$ and $Z = Z + 0$ .

Finding missing values

To find a missing mass number, subtract the known mass numbers on one side from the total on the other side.
To find a missing atomic number, do the same with the bottom numbers, taking care with any $- 1$ for a beta particle.
Once the mass number and charge of an unknown particle are found, they identify it, so a particle of $24{}^{4}_{2}$ is an alpha particle and a particle of $−10{}^{0}_{-1}$ is a beta particle.
Once the atomic number of an unknown nucleus is found, it identifies the element.

Example

Identifying an unknown particle

Radon-222 decays to polonium-218 and emits a particle $X\text{X}$ .

{}^{222}_{86}\text{Rn} \rightarrow {}^{218}_{84}\text{Po} + {}^{a}_{b}\text{X}

The mass numbers must balance.

\quad \Rightarrow \quad a = 4

The charges must balance.

\quad \Rightarrow \quad b = 2

A particle with a mass number of $4$ and a charge of $+ 2$ is an alpha particle, $24He{}^{4}_{2}\text{He}$ .

Example

Finding an unknown nucleus

A nucleus of sodium-24, $1124Na{}^{24}_{11}\text{Na}$ , emits a beta particle and forms a nucleus $Y\text{Y}$ .

{}^{24}_{11}\text{Na} \rightarrow {}^{c}_{d}\text{Y} + {}^{0}_{-1}\text{e}

The mass numbers must balance.

\quad \Rightarrow \quad c = 24

The charges must balance, and the beta particle counts as $- 1$ .

\quad \Rightarrow \quad d = 12

The element with atomic number $12$ is magnesium, so $Y\text{Y}$ is $1224Mg{}^{24}_{12}\text{Mg}$ .

Common Mistake

Do not subtract $1$ from the atomic number in beta decay, because the daughter must gain a proton for the charges to balance with the $- 1$ of the beta particle.
Do not balance only the mass numbers, because an equation is only balanced when the charges also match.
Do not change the element symbol in a gamma equation, because gamma emission leaves the nucleus as the same isotope.

Exam technique

Showing your balancing

Write the two totals as a check, for example $238 = 234 + 4$ and $92 = 90 + 2$ , so that the examiner can see the conservation of mass and charge.
When a question asks you to "explain" why an equation is balanced, refer to both the total mass number and the total charge on each side.
Give a missing particle by its name and its full symbol, because the name alone or the symbol alone may not get both marks.

Self review

Which two totals must be the same on both sides of a balanced decay equation?
Why does the daughter nucleus in beta decay have an atomic number one greater than the parent?
What are the missing numbers in $84218Po→baPb+24He{}^{218}_{84}\text{Po} \rightarrow {}^{a}_{b}\text{Pb} + {}^{4}_{2}\text{He}$ ?
What does an asterisk next to a nuclear symbol show?
Why does a gamma ray not change either total in a decay equation?

Conservation of mass and charge

Definition

Balanced decay equation

A decay equation in which the total mass number and the total charge are the same on both sides of the arrow.

A decay equation is balanced when the totals on the two sides of the arrow are the same.
The total mass number on the left must equal the total mass number on the right, because no nucleons are created or destroyed.
The total charge on the left must equal the total charge on the right, because charge is conserved.
The charge of each nucleus is its atomic number, so the bottom numbers must add up to the same total on both sides.
The charge of a beta particle is $- 1$ , so it must be counted as a negative number when the bottom numbers are added.
These two rules let you find any single missing number, nucleus or particle in a decay equation.

Balancing alpha decay

The general equation for alpha decay is $ZAX→Z−2A−4Y+24He{}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\text{He}$ .
The mass numbers balance because $A = (A - 4) + 4$ .
The charges balance because $Z = (Z - 2) + 2$ .
For uranium-235 decaying to thorium-231, the mass numbers balance as $235 = 231 + 4$ and the charges balance as $92 = 90 + 2$ .

Balancing beta decay

The general equation for beta decay is $ZAX→Z+1AY+−10e{}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z+1}\text{Y} + {}^{0}_{-1}\text{e}$ .
The mass numbers balance because $A = A + 0$ .
The charges balance because $Z = (Z + 1) + (- 1)$ .
The daughter nucleus has one more positive charge, and the beta particle carries away one negative charge, so the total charge stays the same.
For thorium-234 decaying to protactinium-234, the mass numbers balance as $234 = 234 + 0$ and the charges balance as $90 = 91 + (- 1)$ .

Balancing gamma and neutron emission

A nucleus with excess energy is often marked with an asterisk, so $ZAX∗{}^{A}_{Z}\text{X}^{*}$ is a nucleus that is about to emit a gamma ray.
The general equation for gamma emission is $ZAX∗→ZAX+00γ{}^{A}_{Z}\text{X}^{*} \rightarrow {}^{A}_{Z}\text{X} + {}^{0}_{0}\gamma$ .
The gamma ray adds $0$ to both totals, so the parent and daughter have the same mass number and the same atomic number.
For example, $4399Tc∗→4399Tc+00γ{}^{99}_{43}\text{Tc}^{*} \rightarrow {}^{99}_{43}\text{Tc} + {}^{0}_{0}\gamma$ shows an excited technetium-99 nucleus losing energy.
The general equation for neutron emission is $ZAX→ZA−1X+01n{}^{A}_{Z}\text{X} \rightarrow {}^{A-1}_{Z}\text{X} + {}^{1}_{0}\text{n}$ , which balances because $A = (A - 1) + 1$ and $Z = Z + 0$ .

Finding missing values

To find a missing mass number, subtract the known mass numbers on one side from the total on the other side.
To find a missing atomic number, do the same with the bottom numbers, taking care with any $- 1$ for a beta particle.
Once the mass number and charge of an unknown particle are found, they identify it, so a particle of $24{}^{4}_{2}$ is an alpha particle and a particle of $−10{}^{0}_{-1}$ is a beta particle.
Once the atomic number of an unknown nucleus is found, it identifies the element.

Example

Identifying an unknown particle

Radon-222 decays to polonium-218 and emits a particle $X\text{X}$ .

{}^{222}_{86}\text{Rn} \rightarrow {}^{218}_{84}\text{Po} + {}^{a}_{b}\text{X}

The mass numbers must balance.

\quad \Rightarrow \quad a = 4

The charges must balance.

\quad \Rightarrow \quad b = 2

A particle with a mass number of $4$ and a charge of $+ 2$ is an alpha particle, $24He{}^{4}_{2}\text{He}$ .

Example

Finding an unknown nucleus

A nucleus of sodium-24, $1124Na{}^{24}_{11}\text{Na}$ , emits a beta particle and forms a nucleus $Y\text{Y}$ .

{}^{24}_{11}\text{Na} \rightarrow {}^{c}_{d}\text{Y} + {}^{0}_{-1}\text{e}

The mass numbers must balance.

\quad \Rightarrow \quad c = 24

The charges must balance, and the beta particle counts as $- 1$ .

\quad \Rightarrow \quad d = 12

The element with atomic number $12$ is magnesium, so $Y\text{Y}$ is $1224Mg{}^{24}_{12}\text{Mg}$ .

Common Mistake

Do not subtract $1$ from the atomic number in beta decay, because the daughter must gain a proton for the charges to balance with the $- 1$ of the beta particle.
Do not balance only the mass numbers, because an equation is only balanced when the charges also match.
Do not change the element symbol in a gamma equation, because gamma emission leaves the nucleus as the same isotope.

Exam technique

Showing your balancing

Write the two totals as a check, for example $238 = 234 + 4$ and $92 = 90 + 2$ , so that the examiner can see the conservation of mass and charge.
When a question asks you to "explain" why an equation is balanced, refer to both the total mass number and the total charge on each side.
Give a missing particle by its name and its full symbol, because the name alone or the symbol alone may not get both marks.

Self review

Which two totals must be the same on both sides of a balanced decay equation?
Why does the daughter nucleus in beta decay have an atomic number one greater than the parent?
What are the missing numbers in $84218Po→baPb+24He{}^{218}_{84}\text{Po} \rightarrow {}^{a}_{b}\text{Pb} + {}^{4}_{2}\text{He}$ ?
What does an asterisk next to a nuclear symbol show?
Why does a gamma ray not change either total in a decay equation?

6.1.7 Balancing mass and charge in decay

Conservation of mass and charge

Balancing alpha decay

Balancing beta decay

Balancing gamma and neutron emission

Finding missing values

Identifying an unknown particle

Finding an unknown nucleus

Showing your balancing

Review 6.1.7 Balancing mass and charge in decay by teaching Genie

6.1.7 Balancing mass and charge in decay

Conservation of mass and charge

Balancing alpha decay

Balancing beta decay

Balancing gamma and neutron emission

Finding missing values

Identifying an unknown particle

Finding an unknown nucleus

Showing your balancing

Review 6.1.7 Balancing mass and charge in decay by teaching Genie

6.1.7 Balancing mass and charge in decay

Conservation of mass and charge

Balancing alpha decay

Balancing beta decay

Balancing gamma and neutron emission

Finding missing values

Identifying an unknown particle

Finding an unknown nucleus

Showing your balancing

Review 6.1.7 Balancing mass and charge in decay by teaching Genie

6.1.7 Balancing mass and charge in decay Revision Guide

6.1.7 Balancing mass and charge in decay

Conservation of mass and charge

Balancing alpha decay

Balancing beta decay

Balancing gamma and neutron emission

Finding missing values

Identifying an unknown particle

Finding an unknown nucleus

Showing your balancing

Review 6.1.7 Balancing mass and charge in decay by teaching Genie

6.1.7 Balancing mass and charge in decay Revision Guide