Why Beta-Minus Decay Increases Atomic Number but Not Mass Number

Carbon-14 contains 6 protons and 8 neutrons. After beta-minus decay, it becomes nitrogen-14, which contains 7 protons and 7 neutrons.

The element has changed, but the mass number is still 14. How can a nucleus gain a proton without gaining another nucleon?

The answer is that a neutron changes into a proton. One neutron disappears, one proton takes its place, and the total number of nucleons stays constant. The emitted beta particle is an electron, not a nucleon, so it does not reduce the mass number.

There is another detail worth keeping separate: an unchanged mass number does not mean an unchanged physical mass. The parent and daughter atoms can have slightly different actual masses, and that difference is connected to the energy released during decay.

01Follow the nucleons

Start with two quantities:

• Atomic number, \(Z\): the number of protons in the nucleus.
• Mass number, \(A\): the total number of protons and neutrons.

If \(N\) represents the number of neutrons, then:

\[ A=Z+N \]

This means:

\[ N=A-Z \]

For carbon-14, the starting nucleus contains:

\[ Z=6,\qquad N=14-6=8 \]

During beta-minus decay, one neutron transforms into a proton.

The changes are therefore:

• Proton number increases by one.
• Neutron number decreases by one.
• Total nucleon number remains unchanged.

For carbon-14, the transformation looks like this:

The important point is that the nucleus does not simply lose a neutron. If it did, the total nucleon number would decrease.

Instead, the neutron changes identity.

Why does the atomic number increase?

Atomic number is determined entirely by the number of protons.

Carbon has 6 protons. Nitrogen has 7.

When a neutron becomes a proton, the nucleus has one additional proton, so it becomes a different element.

This is why beta-minus decay increases atomic number by one:

\[ Z_{\text{daughter}}=Z_{\text{parent}}+1 \]

Why does the mass number stay the same?

Mass number counts protons and neutrons together.

The neutron-to-proton transformation changes the type of nucleon, not the total number of nucleons.

The mass number therefore remains:

\[ A_{\text{daughter}}=A_{\text{parent}} \]

The emitted electron does not count towards \(A\). It has a mass number of zero because it is not a proton or neutron.

The complete transformation inside the nucleus is:

\[ n\rightarrow p+e^-+\overline{\nu}_e \]

Here:

• \(n\) is a neutron.
• \(p\) is a proton.
• \(e^-\) is the emitted electron, or beta-minus particle.
• \(\overline{\nu}_e\) is an electron antineutrino.

The electron is created during the nuclear transformation. It is not an electron that was previously orbiting the atom.

Check both conservation laws

There are two useful checks when writing a beta-minus decay equation.

1. Conservation of charge

A neutron has zero charge. The products have charges of \(+1\), \(-1\), and \(0\).

\[ 0=(+1)+(-1)+0 \]

Total charge is conserved.

2. Conservation of nucleon number

A neutron contributes one nucleon. A proton contributes one nucleon, while the electron and antineutrino contribute none.

\[ 1=1+0+0 \]

Total nucleon number is conserved.

This gives the general beta-minus decay equation:

\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e + {}^{0}_{0}\overline{\nu}_e \]

Notice that the upper numbers balance independently of the lower numbers:

\[ A=A+0+0 \]

\[ Z=(Z+1)-1+0 \]

Both conservation laws are satisfied.

For a fuller explanation of the emitted particles and the role of the antineutrino, see Beta Decay Explained: Nuclear Equations and Emitted Particles.

What would happen if a neutron simply left?

This is an important comparison because it exposes a common mistake.

Suppose a nucleus actually ejects a neutron rather than undergoing beta-minus decay.

The process would be:

\[ {}^{A}_{Z}X \rightarrow {}^{A-1}_{Z}Y + {}^{1}_{0}n \]

In this case:

• The proton number stays the same.
• The neutron number decreases by one.
• The mass number decreases by one.
• The element stays the same.

Compare the two processes:

The distinction is simple: losing a neutron reduces the mass number, while converting a neutron into a proton does not.

If you are unsure which process is occurring, check whether the proton number changes. That tells you whether the nucleus has undergone a transformation between nucleons or simply lost a neutron.

02The historical clue: how scientists worked out what was happening

The explanation becomes more interesting when you consider what scientists knew before the neutron was discovered.

Radioactive decay was already known to change one element into another. But the internal structure of the nucleus was still being worked out.

The first clue: the displacement law

In the early twentieth century, Frederick Soddy and other researchers established patterns in radioactive transformations.

Soddy’s displacement law described how beta emission moves an element one position higher in the periodic table. Alpha emission moves it two positions lower.

For beta emission, this meant that the atomic number increased by one.

The experimental pattern was clear, but the internal explanation was not.

How could an element gain one unit of positive nuclear charge without gaining another unit of nuclear mass?

At the time, the neutron had not yet been established as a nuclear particle. Scientists needed evidence for a neutral particle that could contribute mass without contributing electrical charge.

Chadwick’s experiment: looking for a neutral particle

In 1932, physicist James Chadwick investigated mysterious radiation produced when alpha particles struck beryllium.

The experimental arrangement was important.

A radioactive polonium source emitted alpha particles towards a beryllium target. The beryllium produced a highly penetrating, electrically neutral radiation.

This radiation then travelled towards a piece of paraffin wax.

Paraffin contains many hydrogen atoms. Each hydrogen nucleus is a single proton, making it useful for investigating collisions with incoming radiation.

When the radiation struck the paraffin, fast-moving protons were knocked out of the material. These recoiling protons could then be detected through their ionisation and tracks in suitable detectors.

What should the experiment show?

There were two competing explanations for the radiation.

One possibility was that it consisted of high-energy gamma photons.

The other was that it consisted of neutral particles with substantial mass.

If the radiation consisted of photons, the observed proton recoil behaviour should be consistent with energy and momentum transferred by photon collisions.

However, the measured recoil behaviour was difficult to reconcile with the gamma-ray explanation.

Chadwick extended the investigation using different target materials, including helium, nitrogen, and lithium. He also used different detection methods to examine the recoil particles.

The results supported a different interpretation: the incoming radiation consisted of neutral particles with masses close to that of protons.

These particles were neutrons.

The discovery of the neutron in 1932 provided a much clearer model of the nucleus. It contained two types of nucleons:

• Protons, which contribute positive charge.
• Neutrons, which contribute mass but have no electrical charge.

An important historical distinction is that Chadwick’s experiment established the neutron’s existence. It did not directly demonstrate beta-minus decay itself. The neutron’s discovery made the later nuclear explanation much more coherent.

The combination of the observed radioactive displacement and the neutron-proton model explains the original puzzle: a neutron can change into a proton without changing the total number of nucleons.

For historical background, see the American Physical Society’s account of Chadwick’s discovery.

03Mass number is not the same as physical mass

Now comes a distinction that matters when you start calculating nuclear energy.

If beta-minus decay leaves the mass number unchanged, does that mean the parent and daughter have exactly the same mass?

No.

Mass number is a count of nucleons. Physical mass is a measured quantity.

These are related, but they are not interchangeable.

Consider four different quantities:

A mass number must be a whole number because it counts particles.

Nuclear and atomic masses are measured quantities. They are usually expressed in unified atomic mass units, \(\mathrm{u}\), and generally contain decimal places.

For example, carbon-14 has a mass number of 14, but its neutral atomic mass is approximately \(14.003242\ \mathrm{u}\).

The two numbers describe different properties.

What does binding energy have to do with mass?

Imagine taking a collection of free protons and neutrons and bringing them together to form a nucleus.

The resulting bound nucleus has less rest mass than the sum of the masses of its separated nucleons.

The difference is called the mass defect.

It corresponds to the energy released when the nucleus forms. The same amount of energy would be required to separate the nucleus completely into free nucleons.

This energy is called the nuclear binding energy.

For a nucleus containing \(Z\) protons and \(N\) neutrons, its total binding energy is:

\[ B=\left[Zm_p+Nm_n-M_{\text{nuc}}\right]c^2 \]

where:

• \(m_p\) is the rest mass of a free proton.
• \(m_n\) is the rest mass of a free neutron.
• \(M_{\text{nuc}}\) is the actual nuclear mass.
• \(c\) is the speed of light.

A larger binding energy means more energy is required to separate the nucleus into free nucleons.

Importantly, binding energy is not the same as the energy released in a particular radioactive decay.

Nuclear mass versus atomic mass

There is another distinction to keep straight.

A nuclear mass includes the nucleus but excludes the surrounding electrons.

An atomic mass includes the nucleus and the electrons of a neutral atom.

This matters because beta-minus decay changes the nuclear charge by one. The daughter element has one more proton than the parent.

When using nuclear masses to calculate the energy released in beta-minus decay, the emitted electron must be included explicitly:

\[ Q_{\beta^-} = \left[ M_{\text{nuc},P} - M_{\text{nuc},D} - m_e \right]c^2 \]

Here:

• \(M_{\text{nuc},P}\) is the parent nuclear mass.
• \(M_{\text{nuc},D}\) is the daughter nuclear mass.
• \(m_e\) is the electron rest mass.
• \(Q_{\beta^-}\) is the energy released.

There is a useful alternative.

If the question supplies neutral atomic masses rather than bare nuclear masses, the beta-minus energy can be calculated using:

\[ Q_{\beta^-} \approx \left[ M_{\text{atom},P} - M_{\text{atom},D} \right]c^2 \]

You do not subtract the electron mass again.

The extra electron mass included in the neutral daughter atomic mass accounts for the emitted beta electron in this calculation. Small differences in electron binding energies are neglected at the usual HSC calculation precision.

Always identify which type of mass the question supplies before choosing your formula.

Binding energy versus released energy

The decay energy, \(Q\), describes the decrease in total rest-mass energy between the initial and final states.

Using nuclear masses:

\[ Q=(M_{\text{initial}}-M_{\text{final}})c^2 \]

A positive \(Q\) means energy is released.

A negative \(Q\) means the proposed decay cannot occur spontaneously to that final state without an additional energy source.

For beta-minus decay, the relationship between binding energy and released energy can be written as:

\[ Q_{\beta^-} = (m_n-m_p-m_e)c^2 + (B_D-B_P) \]

Here, \(B_P\) and \(B_D\) are the total binding energies of the parent and daughter nuclei.

This equation reveals something important.

The released energy depends on two things:

1. The difference between the rest masses of a free neutron and the products of its transformation.
2. The change in nuclear binding energy between the parent and daughter.

Therefore, you cannot automatically equate the energy released in beta decay with the increase in binding energy.

The daughter can even have a slightly lower total binding energy while the decay still releases energy. The difference in free-particle rest masses also contributes.

Finally, the released energy does not all go into the beta particle. It is shared between the beta electron, the antineutrino, and the recoiling daughter nucleus. If the daughter is initially excited, some energy may also be released later as gamma radiation.

06Questions and solutions

Question 1

A sulfur-35 nucleus undergoes beta-minus decay.

Determine:

a) The daughter element and its mass number.

b) The number of protons and neutrons in both the parent and daughter nuclei.

c) The complete balanced nuclear equation.

Solution 1

The daughter is chlorine-35, containing 17 protons and 18 neutrons.

Sulfur has atomic number 16, so the parent contains:

\[ Z=16,\qquad N=35-16=19 \]

Beta-minus decay increases the proton number by one:

\[ Z_D=16+1=17 \]

Atomic number 17 corresponds to chlorine.

The daughter neutron number is:

\[ N_D=35-17=18 \]

The balanced equation is:

\[ {}^{35}_{16}\mathrm{S} \rightarrow {}^{35}_{17}\mathrm{Cl} + {}^{0}_{-1}e + {}^{0}_{0}\overline{\nu}_e \]

The mass number remains 35 because one neutron becomes one proton.

Checking charge:

\[ 16=17-1 \]

Checking nucleon number:

\[ 35=35 \]

Both conservation laws are satisfied.

Question 2

A neutral potassium-41 atom undergoes beta-minus decay.

Immediately after the nuclear transformation, assume that its original 19 atomic electrons remain associated with the daughter atom, while the emitted beta electron has escaped.

a) Identify the daughter nucleus.

b) Determine the electrical charge of the daughter atom immediately after decay.

c) Show that total electrical charge is conserved when the emitted beta electron is included.

d) The daughter later captures an electron from its surroundings. Explain what changes and what remains unchanged.

Solution 2

The daughter nucleus is calcium-41, and immediately after decay the daughter atom is a singly positive ion.

Part a: Identify the daughter.

Potassium has atomic number 19.

Therefore:

\[ Z_D=19+1=20 \]

Atomic number 20 corresponds to calcium.

The mass number remains 41:

\[ {}^{41}_{19}\mathrm{K} \rightarrow {}^{41}_{20}\mathrm{Ca} + {}^{0}_{-1}e + {}^{0}_{0}\overline{\nu}_e \]

Part b: Determine the daughter’s electrical charge.

The daughter nucleus contains 20 protons.

However, the original atom had only 19 orbital electrons, and the emitted beta electron has left.

The daughter therefore has:

• 20 positive proton charges.
• 19 negative electron charges.

Its net charge is:

\[ +20-19=+1 \]

It is a calcium-41 positive ion.

Part c: Check total charge conservation.

Before decay, the neutral potassium atom has zero net charge.

After decay, the products have:

• Daughter ion: \(+1\).
• Beta electron: \(-1\).
• Antineutrino: \(0\).

Therefore:

\[ 0=(+1)+(-1)+0 \]

Total charge is conserved.

Part d: Consider electron capture from the surroundings.

When the calcium ion captures an additional electron, it becomes electrically neutral.

However, the surrounding material has lost an electron and is left with an additional positive charge.

The total charge of the complete system remains unchanged.

Most importantly, capturing an electron does not change the nuclear composition. Calcium still has 20 protons and 21 neutrons, so its mass number remains 41.

The trap is confusing the electrical charge of an atom with its atomic number. Atomic number counts protons, not electrons.

Question 3

A hypothetical nucleus undergoes beta-minus decay.

Both the parent and daughter nuclei have mass number 40. Their measured bare nuclear masses are:

Use \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).

a) Calculate the energy released.

b) Explain why the electron rest mass must be included in the calculation.

c) Explain how the physical mass can decrease even though the mass number remains 40.

The supplied masses are illustrative values for this hypothetical transition, not tabulated isotope data.

Solution 3

The decay releases approximately \(0.420\ \mathrm{MeV}\).

The important detail is that the supplied values are nuclear masses, not neutral atomic masses.

Part a: Calculate the released energy.

For beta-minus decay using nuclear masses:

\[ Q=(M_P-M_D-m_e)c^2 \]

Substitute the supplied values:

\[ \begin{aligned} \Delta m &=39.962000\\ &\quad-39.961000\\ &\quad-0.0005486\\ &=0.0004514\ \mathrm{u} \end{aligned} \]

Convert the mass difference into energy:

\[ \begin{aligned} Q&=0.0004514\times931.5\\ &=0.4205\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \boxed{Q\approx0.420\ \mathrm{MeV}} \]

Part b: Explain the electron-mass correction.

The nuclear mass measurements describe only the nuclei.

The final products include a daughter nucleus and an emitted electron. The electron therefore needs to be included in the final rest-mass calculation.

This is why its mass is subtracted when using bare nuclear masses.

If the question had supplied neutral atomic masses instead, the electron correction would already be accounted for in the mass difference.

Subtracting it again would give an incorrect result.

Part c: Explain the unchanged mass number.

The mass number counts nucleons, not physical mass.

The parent and daughter both contain 40 nucleons.

One neutron has become one proton, so:

\[ \Delta Z=+1,\qquad \Delta N=-1 \]

Therefore:

\[ \Delta A=+1-1=0 \]

The physical rest mass decreases because the total rest mass of the initial system exceeds that of the final products.

The energy equivalent of this decrease is released as decay energy.

Question 4

A parent atom undergoes beta-minus decay to an excited daughter nuclear state.

The following illustrative data are supplied:

• Difference between the parent and ground-state daughter neutral atomic masses: \(0.000700\ \mathrm{u}\).
• Daughter excitation energy: \(0.200\ \mathrm{MeV}\).
• Measured beta electron kinetic energy in one decay: \(0.100\ \mathrm{MeV}\).

Use:

\[ 1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV} \]

Assume the daughter subsequently emits one gamma photon carrying the full excitation energy. Neglect daughter recoil and the antineutrino rest mass.

a) Calculate the total energy available for decay to the daughter ground state.

b) Calculate the energy available during beta decay to the excited daughter state.

c) Determine the antineutrino energy for the measured event.

d) Explain why the gamma energy must not be counted twice.

Solution 4

The antineutrino carries approximately \(0.352\ \mathrm{MeV}\) in this event.

The key is to distinguish the energy available for the complete decay from the smaller amount available to the beta particles when the daughter is formed in an excited state.

Part a: Calculate the total available energy.

The supplied masses are neutral atomic masses, so:

\[ Q_{\text{ground}}=(M_P-M_D)c^2 \]

Substitute:

\[ \begin{aligned} Q_{\text{ground}} &=0.000700\times931.5\\ &=0.65205\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ Q_{\text{ground}}\approx0.652\ \mathrm{MeV} \]

This is the total energy released when the daughter reaches its ground state.

Part b: Account for the excited state.

The daughter is initially \(0.200\ \mathrm{MeV}\) above its ground state.

That energy must be reserved as nuclear excitation energy.

The energy available to the beta electron and antineutrino during the initial decay is:

\[ Q_{\text{excited}} = Q_{\text{ground}}-E^* \]

Substitute:

\[ Q_{\text{excited}}=0.652-0.200 \]

\[ \boxed{Q_{\text{excited}}=0.452\ \mathrm{MeV}} \]

Part c: Calculate the antineutrino energy.

During the beta-decay stage, the available energy is shared between the beta electron, antineutrino, and daughter recoil.

Recoil is negligible under the stated assumption.

Therefore:

\[ Q_{\text{excited}}=K_e+E_{\overline{\nu}} \]

Rearrange:

\[ E_{\overline{\nu}} = Q_{\text{excited}}-K_e \]

\[ =0.452-0.100 \]

\[ \boxed{E_{\overline{\nu}}=0.352\ \mathrm{MeV}} \]

Part d: Explain the gamma-ray energy.

The gamma photon is emitted when the excited daughter moves down to its ground state.

It carries away the \(0.200\ \mathrm{MeV}\) of excitation energy that was reserved during the original beta decay.

The complete energy balance is:

\[ \begin{aligned} Q_{\text{ground}} &=K_e+E_{\overline{\nu}}+E_\gamma\\ &=0.100+0.352+0.200\\ &=0.652\ \mathrm{MeV} \end{aligned} \]

The gamma energy must not be added to the initial energy release a second time. It is already part of the original mass-energy difference.

The tempting mistake is to treat the beta decay and gamma emission as two independent sources of energy. They are two stages of the same energy-releasing process.

Question 5

A hypothetical nucleus contains 20 protons and 24 neutrons. It undergoes beta-minus decay, producing a daughter nucleus with 21 protons and 23 neutrons.

The following total nuclear binding energies are supplied:

Use the following rest energies:

\[ m_nc^2=939.565\ \mathrm{MeV} \]

\[ m_pc^2=938.272\ \mathrm{MeV} \]

\[ m_ec^2=0.511\ \mathrm{MeV} \]

A student argues:

“The daughter has a smaller binding energy, so forming it must require energy. Beta-minus decay cannot release energy in this situation.”

a) Calculate the energy released when the parent decays to the daughter ground state.

b) A second daughter state lies \(0.200\ \mathrm{MeV}\) above the ground state. Determine whether beta-minus decay to this excited state is energetically possible.

c) Explain why the student’s original argument is incorrect. Your explanation must distinguish nucleon number, nuclear binding energy, and total rest-mass energy.

Assume the antineutrino rest mass is negligible, the supplied binding energies are total energies rather than binding energies per nucleon, and there is no external energy input.

Solution 5

The ground-state decay releases \(0.182\ \mathrm{MeV}\), despite the daughter having a smaller total binding energy.

However, decay to the excited state is energetically forbidden because the excitation energy exceeds the available decay energy.

The key is that beta-minus decay changes both the nucleon identities and the nuclear binding energy. Looking at binding energy alone is not enough.

Part a: Calculate the ground-state decay energy.

First, write the transformation:

\[ n\rightarrow p+e^-+\overline{\nu}_e \]

The energy released by converting a free neutron into a proton and electron is:

\[ Q_{\text{free}} = (m_n-m_p-m_e)c^2 \]

Substitute:

\[ \begin{aligned} Q_{\text{free}} &=939.565-938.272-0.511\\ &=0.782\ \mathrm{MeV} \end{aligned} \]

This is the rest-energy difference for a free neutron undergoing beta-minus decay.

Now consider the change in nuclear binding energy:

\[ \Delta B=B_D-B_P \]

\[ \Delta B=319.4-320.0 \]

\[ \Delta B=-0.600\ \mathrm{MeV} \]

The daughter has \(0.600\ \mathrm{MeV}\) less total binding energy than the parent.

The complete decay energy is:

\[ Q_{\beta^-} = Q_{\text{free}}+\Delta B \]

Therefore:

\[ \begin{aligned} Q_{\beta^-} &=0.782-0.600\\ &=0.182\ \mathrm{MeV} \end{aligned} \]

Hence:

\[ \boxed{Q_{\beta^-}=+0.182\ \mathrm{MeV}} \]

The positive result means the ground-state decay can release energy.

Part b: Test the excited state.

The excited daughter requires an additional \(0.200\ \mathrm{MeV}\).

The energy available for decay to that state is:

\[ Q_{\text{excited}} = Q_{\text{ground}}-E^* \]

Substitute:

\[ Q_{\text{excited}}=0.182-0.200 \]

\[ \boxed{Q_{\text{excited}}=-0.018\ \mathrm{MeV}} \]

The result is negative.

Therefore, beta-minus decay to this excited state is energetically forbidden under the stated assumptions.

The ground-state transition is energetically possible, but the excited-state transition is not.

Part c: Explain the student’s mistake.

The student considered only the change in binding energy.

That misses an important contribution: the neutron being converted has a greater rest mass than the proton and electron produced from it.

The free-particle transformation releases \(0.782\ \mathrm{MeV}\).

Although the daughter loses \(0.600\ \mathrm{MeV}\) of binding energy, the free-particle mass difference provides more than enough energy to compensate.

The net result is a release of \(0.182\ \mathrm{MeV}\).

Now check the nucleon numbers.

The parent contains:

\[ A_P=20+24=44 \]

The daughter contains:

\[ A_D=21+23=44 \]

Therefore:

\[ \Delta A=0 \]

The proton number increases by one, the neutron number decreases by one, and the mass number remains unchanged.

The nuclear rest mass, however, decreases because the complete final system has less rest-mass energy than the initial system.

This is the crucial distinction:

• Mass number tells you how many nucleons are present.
• Binding energy tells you how much energy is needed to separate the nucleons.
• Released energy depends on the difference in total rest-mass energy between the initial and final states.

The final complication is the excited state. Even though the ground-state transition releases energy, an excited state can require more energy than the decay makes available.

That is why a positive ground-state decay energy does not automatically mean every possible daughter state can be produced.

07Where this leads next

Once you can distinguish nucleon number from physical mass, the next step is calculating whether a proposed nuclear transformation releases energy.

That requires choosing the correct mass convention, calculating the mass difference, and applying \(E=mc^2\).

For more practice with that calculation, continue with How to Calculate Energy Released from a Nuclear Reaction’s Mass Change.