How to Identify a Missing Particle in a Nuclear Equation

An alpha source hit a piece of beryllium, and an invisible radiation beam emerged. When that radiation struck paraffin wax, protons flew out at high speed. What had left the beryllium: gamma radiation, or a previously unknown particle?

This is the problem behind identifying a missing particle in a nuclear equation. The reliable method is to calculate the missing nucleon number, \(A\), and charge number, \(Z\), separately. These usually identify the particle, although there is an important exception: some different particles have identical \(A\) and \(Z\) values.

The neutron’s discovery provides a useful starting point because it shows how physicists used conservation laws and experimental evidence together to identify something they could not detect directly.

01The experiment that revealed the neutron

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

The experimental arrangement had several parts:

1. A polonium source emitted alpha particles.
2. The alpha particles travelled towards a beryllium target.
3. The beryllium produced highly penetrating radiation.
4. The radiation travelled towards a thin sheet of paraffin wax, which contains many hydrogen nuclei.
5. Protons knocked out of the paraffin entered an ionisation chamber, where their presence could be detected.
6. Measurements of the recoil particles’ tracks and ranges allowed their speeds to be estimated.

The initial interpretation was that the penetrating radiation consisted of high-energy gamma photons.

But there was a problem.

What should happen if the radiation consists of gamma photons?

Gamma photons can transfer momentum to protons, so observing recoiling protons does not automatically rule them out. However, the gamma-ray explanation must also account for the measured recoil speeds.

Chadwick investigated this by comparing the recoil of hydrogen nuclei (protons) with the recoil of heavier nitrogen nuclei. He also used different experimental arrangements, including expansion chambers, to observe recoil tracks from different target materials.

The results supported a different explanation: the radiation consisted of neutral particles with a mass approximately equal to that of a proton.

The particles were also highly penetrating, indicating that they carried no significant electric charge.

These two observations pointed towards a new particle: the neutron.

A historical scale estimate

Chadwick used the maximum recoil speeds of hydrogen and nitrogen nuclei to estimate the mass of the incoming particle.

In a simplified model of head-on elastic collisions, the relationship was:

\[ \frac{U_H}{U_N}=\frac{M+14}{M+1} \]

Here:

• \(U_H\) is the maximum recoil speed of a hydrogen nucleus.
• \(U_N\) is the maximum recoil speed of a nitrogen nucleus.
• \(M\) is the incoming particle’s mass, measured in proton-mass units.
• The value 14 represents the approximate mass of a nitrogen nucleus in proton-mass units.

Using the approximate speeds reported by Chadwick:

\[ U_H=3.7\times10^7\ \mathrm{m\,s^{-1}} \]

\[ U_N=4.7\times10^6\ \mathrm{m\,s^{-1}} \]

Substituting:

\[ \frac{3.7\times10^7}{4.7\times10^6} = \frac{M+14}{M+1} \]

\[ 7.87=\frac{M+14}{M+1} \]

Rearranging:

\[ 7.87M+7.87=M+14 \]

\[ M=\frac{14-7.87}{7.87-1} \]

\[ M\approx0.89 \]

The incoming particle therefore had approximately 0.89 times the mass of a proton.

This was strong evidence for a neutral particle with a mass close to that of a proton.

The calculation uses a simplified collision model, not an equation you need to memorise for the HSC. Its importance is the reasoning: experimental measurements helped identify a particle that could not be detected directly.

Chadwick’s work is described in his 1935 Nobel Lecture, The Neutron and Its Properties.

The nuclear equation confirms the result

The reaction between beryllium-9 and an alpha particle can be represented as:

\[ {}^{9}_{4}\mathrm{Be} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{12}_{6}\mathrm{C} + {}^{1}_{0}\mathrm{n} \]

Check the nucleon numbers:

\[ 9+4=12+1 \]

\[ 13=13 \]

Now check the charge numbers:

\[ 4+2=6+0 \]

\[ 6=6 \]

The missing particle has:

• Nucleon number \(A=1\).
• Charge number \(Z=0\).

That combination identifies a neutron.

The experiment established the physical existence and properties of the neutron. The nuclear equation provides a second, independent way to check that identification.

This brings us to a method you can apply to unfamiliar nuclear reactions without memorising every possible reaction.

02The two quantities you must conserve

Consider the general nuclear notation:

\[ {}^{A}_{Z}\mathrm{X} \]

The two numbers tell you different things.

For example:

\[ {}^{23}_{11}\mathrm{Na} \]

This sodium nucleus contains:

• 11 protons.
• \(23-11=12\) neutrons.
• 23 nucleons altogether.

In a nuclear equation, two quantities are conserved:

1. Total nucleon number.
2. Total electric charge.

The second quantity is represented by \(Z\) in nuclear notation. For a nucleus, it is the atomic number. For an emitted particle, it represents the particle’s charge in units of the elementary charge.

Think of these as two separate scoreboards. Every reaction must balance both, but balancing one does not guarantee that the other balances.

Recognising common particles

Before identifying an unknown particle, become familiar with the combinations of \(A\) and \(Z\) that common nuclear particles carry.

Notice that electrons, positrons, gamma photons, and neutrinos all have a nucleon number of zero.

This does not mean that they all have zero physical mass. It means that they are not made up of nucleons.

The table also reveals an important limitation: a gamma photon and an antineutrino both have \(A=0\) and \(Z=0\). Conservation of nucleon number and charge alone cannot distinguish them.

We will return to this later.

A reliable method for identifying the missing particle

Suppose a nuclear equation contains one unknown particle, represented by \(X\).

Write it as:

\[ \text{Reactants}\rightarrow\text{Products}+X \]

Use the following process.

Step 1: Balance the nucleon numbers.

Add every \(A\) value on the left-hand side and every \(A\) value on the right-hand side.

Set the totals equal and solve for \(A_X\).

Step 2: Balance the charge numbers.

Repeat the process using the lower numbers, \(Z\).

Set the totals equal and solve for \(Z_X\).

Step 3: Identify the particle.

Compare the calculated values with the particle table.

For example:

• \(A=1,\ Z=0\): neutron.
• \(A=1,\ Z=1\): proton.
• \(A=4,\ Z=2\): alpha particle.
• \(A=0,\ Z=-1\): electron.
• \(A=0,\ Z=+1\): positron.

Step 4: Check the completed equation.

Substitute the particle into the original equation and verify that both totals balance.

Do not guess the particle from the reaction type before doing the calculations. The numbers should lead you to the answer.

03Mass number is not the same as nuclear mass

There is another distinction worth understanding before you start using mass data in nuclear equations.

A common mistake is to assume that if the nucleon numbers balance, the actual masses must also balance.

They do not necessarily balance.

The mass number \(A\) is simply a count of nucleons. It has no units.

Nuclear mass, on the other hand, is the actual rest mass of a nucleus. It can be measured in kilograms or atomic mass units (\(\mathrm{u}\)).

There are three related ideas to keep separate.

Nuclear mass versus atomic mass

A neutral atom contains a nucleus surrounded by electrons.

Its atomic mass is approximately:

\[ m_{\text{atom}} = m_{\text{nucleus}}+Zm_e \]

where:

• \(m_{\text{atom}}\) is the mass of the neutral atom.
• \(m_{\text{nucleus}}\) is the nuclear mass.
• \(Z\) is the number of electrons in the neutral atom.
• \(m_e\) is the mass of one electron.

The small energy associated with the electrons’ binding to the atom is neglected in this approximation.

A nuclear mass does not include the surrounding electrons. An atomic mass does.

This distinction becomes important when you calculate nuclear binding energy or the energy released in a reaction.

What is mass defect?

Imagine taking a helium-4 nucleus apart completely.

You would obtain:

• Two free protons.
• Two free neutrons.

The combined mass of those four separate nucleons is greater than the mass of the original bound helium-4 nucleus.

The difference is called the mass defect.

For a nucleus containing \(Z\) protons and \(A-Z\) neutrons:

\[ \Delta m_{\text{defect}} = Zm_p+(A-Z)m_n-m_{\text{nucleus}} \]

where:

• \(m_p\) is the mass of a free proton.
• \(m_n\) is the mass of a free neutron.
• \(m_{\text{nucleus}}\) is the actual nuclear mass.

The mass defect is associated with the energy released when the nucleus forms.

This energy is called the nuclear binding energy:

\[ E_b=\Delta m_{\text{defect}}c^2 \]

Here, \(c\) is the speed of light in a vacuum.

A nucleus with a larger binding energy requires more energy to separate completely into its individual protons and neutrons.

If you are given atomic masses rather than nuclear masses, you can use a slightly different expression. For a neutral atom:

\[ \Delta m_{\text{defect}} \approx Zm({}^{1}_{1}\mathrm{H}) + (A-Z)m_n - m_{\text{atom}} \]

Here, \(m({}^{1}_{1}\mathrm{H})\) is the mass of a neutral hydrogen atom, not a bare proton.

Using hydrogen atom masses allows the electron masses to cancel. This is important because the mass of the helium atom already includes its electrons.

For a fuller explanation, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

Binding energy versus energy released

These two quantities are related, but they describe different things.

Binding energy is a property of a nucleus. It describes the energy required to separate that nucleus into free nucleons.

Energy released, usually represented by \(Q\), describes the net energy change during a nuclear reaction.

The reaction energy is:

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

Here:

• \(m_{\text{initial}}\) is the total initial rest mass.
• \(m_{\text{final}}\) is the total final rest mass.
• \(Q\) is the energy released or absorbed.

The sign convention is:

• \(Q>0\): energy is released.
• \(Q<0\): energy must be supplied.

When \(Q\) is positive, the products have a smaller combined rest mass than the reactants. The difference appears as energy, usually in the kinetic energy of the products or as radiation.

For nuclear masses expressed in atomic mass units:

\[ 1\ \mathrm{u}\,c^2\approx931.5\ \mathrm{MeV} \]

Therefore:

\[ Q(\mathrm{MeV}) = \Delta m(\mathrm{u})\times931.5 \]

Remember that balancing \(A\) and \(Z\) does not mean the total rest mass is unchanged. Nuclear reactions conserve total energy, including rest energy and kinetic energy.

A final mass-related trap: mixing atomic and nuclear masses

Suppose a question gives you the atomic mass of helium-4 but asks for its nuclear binding energy.

You cannot simply substitute the atomic mass into a formula containing bare proton masses and expect the electrons to disappear.

Your mass data must be consistent.

You can either:

• Use nuclear masses throughout, including bare proton and neutron masses.
• Use neutral atomic masses with the hydrogen-atom formula introduced above.

For beta-minus decay, there is another useful precision point. If you calculate the released energy using nuclear masses, you must account for the mass of the emitted electron. If you use neutral atomic masses, the difference in electron numbers is already included in the parent-to-daughter mass difference, to a very good approximation. Do not subtract the electron mass twice.

For a more detailed treatment of the particle transformations involved, read Beta Decay Explained: Nuclear Equations and Emitted Particles.

04Questions and solutions

The following questions move from straightforward particle identification to problems involving reverse reactions, mass calculations, and competing interpretations of experimental evidence.

Question 1

An incoming alpha particle strikes a nitrogen-14 nucleus, producing an oxygen-17 nucleus and an unknown particle \(X\).

\[ {}^{14}_{7}\mathrm{N} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{17}_{8}\mathrm{O} + X \]

Identify \(X\) using conservation of nucleon number and electric charge. Write its complete nuclear notation.

Solution 1

The missing particle is a proton, \({}^{1}_{1}\mathrm{p}\).

First, balance the nucleon numbers:

\[ 14+4=17+A_X \]

\[ 18=17+A_X \]

Therefore:

\[ A_X=1 \]

Next, balance the charge numbers:

\[ 7+2=8+Z_X \]

\[ 9=8+Z_X \]

Therefore:

\[ Z_X=1 \]

A particle with \(A=1\) and \(Z=1\) is a proton.

The completed equation is:

\[ {}^{14}_{7}\mathrm{N} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{17}_{8}\mathrm{O} + {}^{1}_{1}\mathrm{p} \]

The tempting answer is a neutron because the missing particle has one nucleon. However, the charge calculation changes the result. A neutron has \(Z=0\), whereas the missing particle must have \(Z=1\).

Both conservation laws are needed.

Question 2

A neutron strikes an aluminium-27 nucleus, producing sodium-24 and an unknown particle \(X\).

\[ {}^{27}_{13}\mathrm{Al} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{24}_{11}\mathrm{Na} + X \]

Identify the missing particle and explain why balancing only the nucleon numbers would be insufficient.

Solution 2

The missing particle is an alpha particle, \({}^{4}_{2}\mathrm{He}\).

First, balance the nucleon numbers:

\[ 27+1=24+A_X \]

\[ 28=24+A_X \]

Therefore:

\[ A_X=4 \]

Now balance the charge numbers:

\[ 13+0=11+Z_X \]

\[ 13=11+Z_X \]

Therefore:

\[ Z_X=2 \]

A particle with \(A=4\) and \(Z=2\) is a helium-4 nucleus, also known as an alpha particle.

The completed equation is:

\[ {}^{27}_{13}\mathrm{Al} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{24}_{11}\mathrm{Na} + {}^{4}_{2}\mathrm{He} \]

Balancing only \(A\) would tell you that four nucleons are missing. It would not tell you how many of those nucleons are protons.

The charge calculation establishes that two of the four nucleons are protons. This identifies the alpha particle rather than another possible arrangement of four nucleons.

Question 3

A nuclear reaction is represented by:

\[ {}^{A}_{Z}X + {}^{1}_{0}\mathrm{n} \rightarrow {}^{14}_{6}\mathrm{C} + {}^{4}_{2}\mathrm{He} \]

Determine:

(a) The nucleon number \(A\) of the original nucleus.

(b) Its atomic number \(Z\).

(c) The identity of the original nucleus.

(d) The number of neutrons in the original nucleus.

Explain why the incoming neutron must be included when balancing the reaction.

Solution 3

The original nucleus is oxygen-17, \({}^{17}_{8}\mathrm{O}\), containing nine neutrons.

(a) Find the nucleon number.

Conservation of nucleon number gives:

\[ A+1=14+4 \]

\[ A+1=18 \]

Therefore:

\[ A=17 \]

(b) Find the atomic number.

Conservation of charge gives:

\[ Z+0=6+2 \]

Therefore:

\[ Z=8 \]

(c) Identify the original nucleus.

Atomic number 8 corresponds to oxygen.

Combining this with \(A=17\):

\[ X={}^{17}_{8}\mathrm{O} \]

(d) Calculate the number of neutrons.

The number of neutrons is:

\[ N=A-Z \]

\[ N=17-8 \]

\[ N=9 \]

The original nucleus therefore contains 8 protons and 9 neutrons.

A tempting mistake is to add the nucleon numbers of the two products and immediately assign \(A=18\) to the original nucleus.

That ignores the incoming neutron. The left-hand side contains both the original nucleus and an additional nucleon, so one must be subtracted to find the original nucleus’s nucleon number.

Question 4

A helium-4 atom has an atomic mass of \(4.002603\ \mathrm{u}\).

Use the following illustrative mass data:

Calculate the total nuclear binding energy of helium-4 in MeV.

Explain why using the mass of a bare proton with the atomic mass of helium-4 would produce an incorrect result.

Use:

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

Solution 4

The total binding energy of helium-4 is approximately \(28.30\ \mathrm{MeV}\).

Helium-4 contains two protons and two neutrons.

Because the supplied helium mass is an atomic mass, we must use the neutral hydrogen-atom mass rather than the bare proton mass.

Step 1: Write the mass-defect equation.

Using atomic masses:

\[ \Delta m_{\text{defect}} = Zm_H+(A-Z)m_n-m_{\text{atom}} \]

For helium-4:

\[ \Delta m_{\text{defect}} = 2m_H+2m_n-m_{\mathrm{He}} \]

Step 2: Substitute the supplied values.

\[ \begin{aligned} \Delta m_{\text{defect}} ={}&2(1.007825)\\ &+2(1.008665)\\ &-4.002603 \end{aligned} \]

\[ \Delta m_{\text{defect}} = 2.015650+2.017330-4.002603 \]

\[ \Delta m_{\text{defect}}=0.030377\ \mathrm{u} \]

Step 3: Convert the mass defect into binding energy.

\[ E_b=\Delta m_{\text{defect}}c^2 \]

\[ E_b=0.030377\times931.5 \]

\[ E_b\approx28.30\ \mathrm{MeV} \]

The total binding energy of helium-4 is therefore approximately:

\[ \boxed{E_b=28.30\ \mathrm{MeV}} \]

This is the energy required to separate the helium-4 nucleus completely into two free protons and two free neutrons.

Why can’t we use the bare proton mass?

The atomic mass of helium-4 includes two electrons.

The bare proton mass does not.

If we use two bare proton masses but retain the atomic mass of helium-4, the electron masses are no longer balanced. This introduces an error of approximately two electron masses, equivalent to about \(1.02\ \mathrm{MeV}\).

Using neutral hydrogen-atom masses on the starting side ensures that the two electron masses cancel.

The important lesson is that the mass values must refer to compatible systems. A small bookkeeping error can create a noticeable error in the calculated binding energy.

Question 5

A radioactive nucleus undergoes beta-minus decay in two stages.

The first stage is:

\[ {}^{A}_{Z}\mathrm{P} \rightarrow {}^{A}_{Z+1}\mathrm{D}^{*} + {}^{0}_{-1}\mathrm{e} + X \]

The excited daughter nucleus then de-excites:

\[ {}^{A}_{Z+1}\mathrm{D}^{*} \rightarrow {}^{A}_{Z+1}\mathrm{D} + {}^{0}_{0}\gamma \]

An experiment produces the following observations:

• A gamma detector records one photon with an energy of \(0.80\ \mathrm{MeV}\) in every decay.
• The beta electrons have a continuous range of kinetic energies rather than one fixed energy.
• After accounting for the electron energy, daughter recoil, and the \(0.80\ \mathrm{MeV}\) photon, the remaining energy varies between \(0.05\ \mathrm{MeV}\) and \(0.85\ \mathrm{MeV}\).
• The detector captures all gamma photons in the relevant energy range.
• For this question, the only proposed identities for \(X\) are a gamma photon and an electron antineutrino.

Determine:

(a) The values of \(A_X\) and \(Z_X\).

(b) Whether conservation of nucleon number and charge alone can distinguish the two proposed particles.

(c) The identity of \(X\), using the experimental observations.

(d) Why the continuously varying electron energy is important evidence, even though the nucleon number and charge already balance.

Solution 5

The missing particle is an electron antineutrino, \({}^{0}_{0}\bar{\nu}_e\).

However, conservation of nucleon number and charge alone cannot establish that identity. The experimental evidence is needed to distinguish it from a gamma photon.

(a) Calculate the missing nucleon number.

The original nucleus has nucleon number \(A\).

The products have nucleon numbers:

\[ A,\quad 0,\quad A_X \]

Therefore:

\[ A=A+0+A_X \]

\[ A_X=0 \]

(b) Calculate the missing charge number.

The original nucleus has charge number \(Z\).

The products have charge numbers:

\[ Z+1,\quad -1,\quad Z_X \]

Therefore:

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

Simplifying:

\[ Z=Z+Z_X \]

\[ Z_X=0 \]

The missing particle therefore has:

\[ \boxed{A_X=0,\qquad Z_X=0} \]

Both a gamma photon and an electron antineutrino have these values.

The nuclear equation cannot distinguish between them using these two conservation laws alone.

(c) Identify the particle using the experimental evidence.

The particle is an electron antineutrino.

The key is that the experiment has already identified the \(0.80\ \mathrm{MeV}\) gamma photon as a separate product of the excited daughter nucleus.

If \(X\) were another gamma photon, the detector would register an additional photon. However, no additional gamma photons are detected.

The remaining energy must therefore be carried by the other proposed neutral particle: the electron antineutrino.

(d) Explain the continuous electron energy distribution.

The parent nucleus has a fixed initial rest energy. The total energy available from the decay is therefore fixed.

However, the beta electrons do not all leave with the same kinetic energy.

This tells us that the energy is being shared between multiple products.

The electron antineutrino carries away a variable amount of energy and momentum. Some decays give more energy to the electron, while others give more to the antineutrino.

The energy balance can be represented as:

\[ Q= E_e+E_{\mathrm{recoil}}+E_\gamma+E_{\bar{\nu}} \]

where:

• \(Q\) is the total energy available from the decay.
• \(E_e\) is the beta electron’s kinetic energy.
• \(E_{\mathrm{recoil}}\) is the daughter nucleus’s recoil kinetic energy.
• \(E_\gamma\) is the gamma photon energy.
• \(E_{\bar{\nu}}\) is the energy carried by the antineutrino.

The gamma photon has a fixed energy of \(0.80\ \mathrm{MeV}\), but the remaining energy is shared differently between the electron and antineutrino in different decays.

The tempting answer is to identify \(X\) as a gamma photon because both \(A\) and \(Z\) are zero.

That answer fails because it ignores the experimental evidence. The measured gamma photon already accounts for the daughter’s de-excitation, and the continuously varying beta energy requires another energy and momentum carrier.

The antineutrino provides that explanation.

This is an important distinction between using conservation laws to identify a particle’s possible properties and using experimental evidence to establish its actual identity.

05Where to go next

Identifying a missing particle is the first part of analysing a nuclear reaction.

Once you know what the reaction produces, the next step is to determine whether energy is released or absorbed. That requires comparing the total rest masses of the reactants and products, then connecting the mass difference to binding energy through \(E=mc^2\).

Keep the two stages separate:

• Use \(A\) and \(Z\) to identify missing particles.
• Use actual nuclear or appropriately handled atomic masses to calculate energy changes.

Course-scope note: NESA’s Physics 11-12 Syllabus (2025) begins implementation for Year 11 in Term 1, 2027. Year 12 transitions in Term 4, 2027, with the first HSC examination under the new syllabus in 2028. The 2017 syllabus continues for existing cohorts through 2026. This resource explains the relevant nuclear reasoning and includes extension-level mass calculations; it does not imply that every calculation is prescribed for every syllabus cohort. Check the NESA Physics syllabus implementation information alongside your school’s course program.