How to Check Charge and Nucleon Conservation in Fission

A fission equation can pass one conservation check and still be wrong. For example, the mass numbers in a proposed reaction might add to 236 on both sides, while the proton numbers add to 92 on the left and 93 on the right.

That reaction is not balanced.

To check charge and nucleon conservation in a fission equation, perform two separate calculations:

1. Add the mass numbers (top numbers) of every particle on each side. The totals must match.
2. Add the atomic numbers (bottom numbers) of every particle on each side. These totals must also match.

Both conditions must be satisfied. Getting the mass numbers right does not guarantee that charge is conserved.

There is another detail worth remembering before you start. If a uranium-235 nucleus absorbs a neutron, does its atomic number increase from 92 to 93?

No. A neutron has no electric charge. It increases the total number of nucleons by one but does not change the number of protons.

That distinction is the foundation of balancing nuclear equations.

01Understanding the numbers in a nuclear equation

Consider the general notation for an isotope:

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

Each part tells you something different.

• \(X\) is the chemical symbol of the element.
• \(A\) is the mass number, also called the nucleon number.
• \(Z\) is the atomic number, which tells you the number of protons.

The number of neutrons is therefore:

\[ N=A-Z \]

For example, uranium-235 has 92 protons and:

\[ N=235-92=143 \]

neutrons.

The two numbers in the isotope notation represent two different conservation requirements.

Notice the last row. The number of neutrons in an individual nucleus does not have to stay constant. Neutrons can be absorbed or emitted during fission.

However, the total number of nucleons must remain constant.

Why the neutron is important

A free neutron is written as:

\[ {}^{1}_{0}\mathrm{n} \]

It has:

• Mass number \(A=1\).
• Atomic number \(Z=0\).

When uranium-235 absorbs a neutron, the resulting excited nucleus can be represented as:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{236}_{92}\mathrm{U}^{*} \]

The asterisk indicates that the uranium-236 nucleus is in an excited energy state.

Notice what happened:

• The mass number increased from 235 to 236.
• The atomic number remained 92.

The neutron contributes one nucleon but no positive charge.

This is why you must never add one to both numbers when a nucleus absorbs a neutron.

02Why conservation became important in the discovery of fission

In 1938, Otto Hahn and Fritz Strassmann were investigating what happened when uranium was bombarded with neutrons.

Their original expectation was that neutron absorption could produce heavier elements. However, their chemical analysis revealed radioactive products behaving like barium, an element with atomic number 56.

This was a problem.

Uranium has an atomic number of 92. How could bombarding it with a neutron produce an element with only 56 protons?

The experiment did not involve watching a nucleus split directly. The researchers had to infer what had happened from the chemical properties and radioactive behaviour of the products.

A simplified representation of their experimental process is:

1. A neutron source irradiated a uranium sample.
2. The irradiated material produced radioactive products.
3. The products were chemically separated.
4. The chemical behaviour and radioactivity of the separated fractions were examined.
5. The identification of barium challenged the expectation that uranium would simply produce heavier elements.

Lise Meitner and Otto Frisch subsequently explained the unexpected result using a model in which the heavy nucleus splits into two smaller nuclei.

They recognised that the original uranium nucleus could divide while conserving its total nuclear charge. If one fragment was barium, with 56 protons, the other fragment needed a total of 36 protons to account for uranium’s original 92.

This interpretation established the basis of our modern understanding of nuclear fission.

The historical reasoning is described in the American Physical Society’s account of the discovery of nuclear fission.

The important lesson for balancing equations is that a fission reaction produces several particles and nuclei. You must account for every one of them.

03How to check nucleon number and charge conservation

Let’s turn that idea into a method you can use in an HSC question.

Consider a general fission reaction:

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

There are two independent checks.

Check 1: Conservation of nucleon number

Add the mass numbers of every particle on the reactant side.

Then add the mass numbers of every particle on the product side.

You need:

\[ \boxed{\sum A_{\text{reactants}}= \sum A_{\text{products}}} \]

For example, if uranium-235 absorbs one neutron:

\[ 235+1=236 \]

The products must therefore contain a total of 236 nucleons.

Remember to count free neutrons. Each free neutron contributes one to the mass-number total.

Check 2: Conservation of electric charge

Now add the atomic numbers on each side.

You need:

\[ \boxed{\sum Z_{\text{reactants}}= \sum Z_{\text{products}}} \]

For a nuclear equation, the atomic number represents the nuclear charge in units of the elementary charge.

For example, uranium has 92 protons, so its nucleus has charge \(+92e\).

A neutron contributes zero charge, while an electron contributes \(-e\).

Common particles and their contributions

This table is particularly useful when a question includes emitted neutrons or gamma radiation.

A gamma photon carries energy, but it contributes nothing to either conservation sum.

04A common mistake: Checking mass number but forgetting charge

Consider this proposed reaction:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{95}_{39}\mathrm{Y} + {}^{1}_{0}\mathrm{n} \]

A student checks the mass numbers:

\[ 235+1=140+95+1 \]

\[ 236=236 \]

They conclude that the equation is balanced.

But they have only completed one check.

Now consider the atomic numbers:

\[ Z_{\text{reactants}}=92+0=92 \]

\[ Z_{\text{products}}=54+39+0=93 \]

The product side has one extra unit of positive charge.

The equation is therefore incorrect.

How could we fix it?

If the mass number of the second fragment must remain 95, its atomic number needs to be 38 rather than 39.

The corrected fragment is strontium-95:

\[ {}^{95}_{38}\mathrm{Sr} \]

The corrected equation becomes:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{95}_{38}\mathrm{Sr} + {}^{1}_{0}\mathrm{n} \]

Now both checks work:

\[ A:\quad 236=140+95+1 \]

\[ Z:\quad 92=54+38 \]

The equation satisfies both conservation requirements.

This illustrates an important distinction:

• Adding a neutron can fix a missing nucleon.
• Changing the isotope can fix an incorrect proton number.
• Adding a gamma photon fixes neither, because it contributes zero to both totals.

Never assume that correcting one side of the equation automatically fixes the other.

05Nuclear mass, atomic mass, mass defect and binding energy

So far, we have dealt with particle counts.

But there is another question that often appears alongside fission equations:

If nucleon number is conserved, why do the products have a different total mass from the reactants?

The answer is that nucleon number and mass are not the same thing.

A mass number is an integer count of nucleons. Actual nuclear masses are measured quantities and do not have to equal the mass number multiplied by one atomic mass unit.

This difference is central to nuclear energy.

Nuclear mass versus atomic mass

First, distinguish these two terms.

Nuclear mass is the rest mass of the nucleus itself. It includes the protons and neutrons but excludes the surrounding electrons.

Atomic mass is the rest mass of a neutral atom. It includes the nucleus and its electrons, with a small reduction due to electronic binding energy.

This matters because nuclear physics data tables frequently provide atomic masses rather than nuclear masses.

For example, a neutral uranium-235 atom contains:

• 92 protons.
• 143 neutrons.
• 92 electrons.

Its atomic mass includes all three groups of particles.

For a fission reaction, atomic masses can often be used directly if the number of electrons cancels between reactants and products.

In our uranium-235 example:

• The uranium atom contains 92 electrons.
• The xenon atom contains 54 electrons.
• The strontium atom contains 38 electrons.

The products therefore contain:

\[ 54+38=92\text{ electrons} \]

The electron contributions cancel.

At HSC calculation accuracy, the small differences in electronic binding energies can be ignored.

If electron numbers do not balance, you must account for the electron masses consistently rather than automatically using atomic masses.

What is mass defect?

Imagine separating a nucleus into all its individual protons and neutrons.

You could calculate the total mass of those particles before they were bound together.

However, the actual mass of the bound nucleus is smaller.

The difference is called the mass defect.

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

\[ \Delta m=Zm_p+Nm_n-m_{\text{nucleus}} \]

where:

• \(\Delta m\) is the mass defect in kilograms or atomic mass units.
• \(m_p\) is the mass of one proton.
• \(m_n\) is the mass of one neutron.
• \(m_{\text{nucleus}}\) is the actual nuclear mass.

The mass defect exists because the formation of a bound nucleus involves a reduction in the system’s rest-mass energy.

What is nuclear binding energy?

The mass defect is directly related to nuclear binding energy through Einstein’s mass-energy equivalence:

\[ \boxed{E=mc^2} \]

For a nucleus:

\[ \boxed{B=\Delta mc^2} \]

Here, \(B\) is the nuclear binding energy, \(\Delta m\) is the mass defect, and \(c\) is the speed of light in a vacuum.

Binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons.

A larger total binding energy means more energy is required to pull the nucleus apart.

Do not confuse binding energy with energy released by fission. Breaking a nucleus apart requires energy. Fission releases energy because the resulting fragments can have a greater combined binding energy than the original nucleus.

This explains why heavy nuclei can release energy through fission.

Their fragments are generally more tightly bound per nucleon than the original heavy nucleus. The total binding energy increases, and the total rest-mass energy decreases.

What is the energy released in fission?

The energy released by an entire nuclear reaction is described by its reaction energy, often called the Q-value.

It is calculated by comparing the total rest mass of all reactants with the total rest mass of all products.

Using the sign convention that positive \(Q\) means energy released:

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

Therefore:

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

If the mass difference is given in atomic mass units, use:

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

where:

\[ \Delta m=m_{\text{initial}}-m_{\text{final}} \]

The conversion factor is approximately \(931.5\text{ MeV}\) per atomic mass unit.

There is an important distinction here:

The mass defect of uranium-235 is not the same quantity as the mass difference of an entire fission reaction.

You must compare every reactant with every product.

06A reliable decision process for HSC questions

When you are given a nuclear fission equation, follow this sequence.

Keep these checks separate.

A reaction can conserve nucleon number and charge while still requiring an energy input. Conservation of charge and nucleon number tells you that the equation satisfies those two bookkeeping requirements. It does not, by itself, establish that the reaction is energetically favourable or likely to occur.

Also remember that nuclear mass and atomic mass are different quantities. Always check what type of mass a question supplies before calculating an energy difference.

07Questions and solutions

The following questions progress from straightforward equation balancing to more demanding problems involving radioactive decay, binding energy, and atomic mass.

Question 1

A plutonium-239 nucleus absorbs a neutron and undergoes fission:

\[ {}^{239}_{94}\mathrm{Pu} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{144}_{56}\mathrm{Ba} + {}^{92}_{38}\mathrm{Sr} + x{}^{1}_{0}\mathrm{n} \]

Determine the number of neutrons released, \(x\), and demonstrate that both nucleon number and charge are conserved.

Solution 1

Four neutrons are released, and both conservation requirements are satisfied.

First, conserve nucleon number:

\[ 239+1=144+92+x \]

\[ 240=236+x \]

Therefore:

\[ \boxed{x=4} \]

Now check charge.

For the reactants:

\[ Z_{\text{reactants}}=94+0=94 \]

For the products:

\[ Z_{\text{products}}=56+38+4(0)=94 \]

Both sides have a total nucleon number of 240 and a total charge number of 94.

The completed equation is:

\[ \boxed{ {}^{239}_{94}\mathrm{Pu} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{144}_{56}\mathrm{Ba} + {}^{92}_{38}\mathrm{Sr} + 4{}^{1}_{0}\mathrm{n} } \]

Remember that the four emitted neutrons contribute four nucleons but no charge.

Question 2

A student proposes the following fission reaction:

\[ {}^{238}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{141}_{54}\mathrm{Xe} + {}^{96}_{39}\mathrm{Y} + 2{}^{1}_{0}\mathrm{n} \]

(a) Determine whether nucleon number is conserved.

(b) Determine whether charge is conserved.

(c) If the mass number of the second fragment must remain 96, identify the isotope that would correct the equation.

(d) Explain why correcting the equation does not automatically establish that the reaction releases energy.

Solution 2

Nucleon number is conserved, but charge is not. The second fragment must be strontium-96 to satisfy both conservation requirements.

(a) Nucleon number

The reactant total is:

\[ A_{\text{reactants}}=238+1=239 \]

The product total is:

\[ A_{\text{products}}=141+96+2=239 \]

Therefore, nucleon number is conserved.

(b) Charge

The reactant charge total is:

\[ Z_{\text{reactants}}=92+0=92 \]

The product charge total is:

\[ Z_{\text{products}}=54+39+0=93 \]

Charge is not conserved.

The equation has one additional unit of positive charge on the product side.

(c) Correcting the isotope

The second fragment needs an atomic number of:

\[ Z=92-54=38 \]

Element 38 is strontium.

Therefore, the corrected isotope is:

\[ {}^{96}_{38}\mathrm{Sr} \]

The corrected equation is:

\[ {}^{238}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{141}_{54}\mathrm{Xe} + {}^{96}_{38}\mathrm{Sr} + 2{}^{1}_{0}\mathrm{n} \]

Both nucleon number and charge are now conserved.

(d) Energetic feasibility

The corrected equation satisfies the two conservation requirements, but this does not prove that it releases energy.

We would also need to compare the total initial and final rest masses.

If the final mass is smaller, the difference can be released as energy. If the final mass is larger, energy must be supplied.

This is why balancing a nuclear equation and calculating its reaction energy are two separate tasks.

Question 3

A uranium-235 nucleus absorbs a neutron and undergoes fission:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^{1}_{0}\mathrm{n} \]

The xenon-140 fragment subsequently undergoes two beta-minus decays:

\[ {}^{140}_{54}\mathrm{Xe} \rightarrow {}^{140}_{55}\mathrm{Cs} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

followed by another beta-minus decay to barium-140.

(a) Write the second beta-minus decay equation.

(b) Write the overall nuclear equation after both beta-minus decays have occurred.

(c) Demonstrate that the overall equation conserves nucleon number and charge.

Solution 3

The second decay produces barium-140, and the overall reaction conserves both nucleon number and charge when the two emitted electrons are included.

(a) Second beta-minus decay

During beta-minus decay, a neutron transforms into a proton, producing an electron and an electron antineutrino.

The second decay is:

\[ {}^{140}_{55}\mathrm{Cs} \rightarrow {}^{140}_{56}\mathrm{Ba} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

The mass number remains 140 because a neutron has become a proton. The atomic number increases by one.

(b) Overall equation

After both beta-minus decays, xenon-140 has become barium-140.

The overall reaction is:

\[ \begin{aligned} {}^{235}_{92}\mathrm{U} +{}^{1}_{0}\mathrm{n} \rightarrow{}^{140}_{56}\mathrm{Ba} +{}^{94}_{38}\mathrm{Sr}\\ +2{}^{1}_{0}\mathrm{n} +2{}^{0}_{-1}\mathrm{e} +2\bar{\nu}_e \end{aligned} \]

Every emitted particle must be included.

(c) Nucleon number

The reactant total is:

\[ A_{\text{reactants}}=235+1=236 \]

The product total is:

\[ A_{\text{products}}=140+94+2=236 \]

Electrons and antineutrinos have zero mass number.

Therefore:

\[ \boxed{A_{\text{reactants}}=A_{\text{products}}} \]

Charge

The reactant total is:

\[ Z_{\text{reactants}}=92 \]

The product total is:

\[ Z_{\text{products}}=56+38-2 \]

\[ Z_{\text{products}}=92 \]

Therefore:

\[ \boxed{Z_{\text{reactants}}=Z_{\text{products}}} \]

The two beta electrons each contribute a charge of \(-1\), balancing the two increases in nuclear charge.

This is a useful example of why you must include emitted particles when checking nuclear equations. Ignoring the beta electrons would incorrectly suggest that charge had increased.

Question 4

A student is given the following rounded binding energies for a fission reaction:

The reaction is:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^{1}_{0}\mathrm{n} \]

The student argues:

“Uranium-235 has a binding energy of 1783.9 MeV, so fission must release 1783.9 MeV.”

Explain why this reasoning is incorrect and calculate the energy released using the supplied binding energies.

Solution 4

The reaction releases approximately 184.6 MeV, not 1783.9 MeV.

The student’s mistake is treating the total binding energy of the original nucleus as the energy released during fission.

However, fission does not completely separate uranium into individual protons and neutrons. It produces two smaller, still-bound nuclei.

We must compare the total binding energy before and after the reaction.

Step 1: Calculate the total binding energy of the products.

The products are xenon-140 and strontium-94, together with two free neutrons.

Free neutrons have zero nuclear binding energy.

Therefore:

\[ \begin{aligned} B_{\text{products}} &=1160.7+807.8\\ &=1968.5\text{ MeV} \end{aligned} \]

Step 2: Calculate the initial total binding energy.

The incoming neutron is free and therefore has zero binding energy.

The initial total binding energy is simply:

\[ B_{\text{initial}}=1783.9\text{ MeV} \]

Step 3: Calculate the increase in total binding energy.

\[ \begin{aligned} \Delta B &=B_{\text{products}}-B_{\text{initial}}\\ &=1968.5-1783.9\\ &=184.6\text{ MeV} \end{aligned} \]

Therefore:

\[ \boxed{Q\approx184.6\text{ MeV}} \]

The products have a greater combined binding energy than the original uranium nucleus and incoming free neutron.

This increase in binding energy corresponds to a reduction in total rest-mass energy.

The result agrees with the mass-difference calculation in the worked example, allowing for rounding.

The key distinction is that the energy released by fission is the increase in total binding energy, not the entire binding energy of the original nucleus.

Question 5

A uranium-235 nucleus absorbs a neutron and undergoes fission:

\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^{1}_{0}\mathrm{n} \]

The xenon-140 fragment then undergoes two successive beta-minus decays, eventually becoming barium-140.

A student writes the overall reaction:

\[ \begin{aligned} {}^{235}_{92}\mathrm{U} +{}^{1}_{0}\mathrm{n} \rightarrow{}^{140}_{56}\mathrm{Ba} +{}^{94}_{38}\mathrm{Sr}\\ +2{}^{1}_{0}\mathrm{n} +2{}^{0}_{-1}\mathrm{e} +2\bar{\nu}_e \end{aligned} \]

They correctly check that nucleon number and charge are conserved.

However, when calculating the total reaction energy, they use neutral atomic masses for uranium-235, barium-140, and strontium-94, and then add the rest masses of two additional electrons to the products because two beta electrons were emitted.

The student argues that this is necessary because the electrons appear explicitly in the nuclear equation.

(a) Explain why the student’s reasoning about electron masses is incorrect.

(b) Determine whether the overall nuclear equation conserves nucleon number and charge.

(c) Explain how the reaction energy should be calculated consistently using either nuclear masses or neutral atomic masses.

(d) If the student adds two electron masses to the final masses after already using neutral atomic masses, determine how this affects their calculated Q-value. Use \(m_e=0.000549\text{ u}\) and \(1\text{ u}c^2=931.5\text{ MeV}\).

Solution 5

The student is double-counting the electron masses. Neutral atomic masses already include the electron contributions, so adding the two beta-electron masses again would underestimate the reaction energy.

This is a subtle distinction between writing a nuclear equation and calculating energy using atomic mass data.

(a) Why the electron masses are double-counted

Start by counting the electrons in the neutral atoms.

The initial uranium-235 atom contains:

\[ 92\text{ electrons} \]

The final neutral atoms contain:

\[ 56+38=94\text{ electrons} \]

There are two additional electrons in the final neutral-atom mass total.

These correspond to the two beta-minus decays.

The neutral atomic masses already include the electron rest masses. Therefore, the difference in electron contributions is already accounted for when comparing the initial and final atomic masses.

Adding another two electron masses would count the same contribution twice.

This is a mass-accounting issue, not a failure of charge conservation.

(b) Check nucleon number and charge

For nucleon number:

\[ A_{\text{initial}}=235+1=236 \]

The final total is:

\[ A_{\text{final}}=140+94+2=236 \]

Electrons and antineutrinos contribute zero nucleons.

Therefore, nucleon number is conserved.

For charge:

\[ Z_{\text{initial}}=92 \]

The final total is:

\[ Z_{\text{final}}=56+38-2 \]

\[ Z_{\text{final}}=92 \]

Charge is also conserved.

(c) Two consistent ways to calculate the reaction energy

There are two valid approaches.

Method 1: Use nuclear masses.

Use the masses of the bare nuclei.

The final products must include the masses of the two emitted beta electrons:

\[ m_{\text{final}}= m_{\mathrm{Ba}}+ m_{\mathrm{Sr}}+ 2m_n+ 2m_e \]

The initial mass is:

\[ m_{\text{initial}}= m_{\mathrm{U}}+m_n \]

The reaction energy is then:

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

The antineutrinos also carry energy, although their rest masses are negligible for this HSC calculation.

Method 2: Use neutral atomic masses.

Use the neutral atomic masses of uranium-235, barium-140, and strontium-94.

The electron contributions are already included in these atomic masses.

Therefore, do not add the two beta-electron masses separately.

The energy calculation must still include the initial neutron and both final free neutrons.

The two methods give consistent results when all particle masses and electron contributions are accounted for correctly.

(d) Calculate the error caused by double-counting

The additional mass incorrectly included is:

\[ \Delta m_{\text{error}}=2m_e \]

Substituting:

\[ \Delta m_{\text{error}} =2(0.000549) \]

\[ \Delta m_{\text{error}}=0.001098\text{ u} \]

Convert this mass into energy:

\[ \begin{aligned} E_{\text{error}} &=0.001098\times931.5\\ &=1.022\text{ MeV} \end{aligned} \]

Therefore, the student’s calculated Q-value would be approximately 1.02 MeV too small.

They have incorrectly increased the final mass by counting the beta-electron rest masses twice.

The deeper lesson is that a nuclear equation and a mass calculation use different conventions for representing electrons. You must be consistent with the type of mass supplied.

This is particularly important when a question combines fission with subsequent radioactive decay.

07What to learn next

Checking nucleon number and charge is the first step in analysing a nuclear reaction. Once the equation is balanced, you can investigate whether the reaction releases or absorbs energy by comparing the total initial and final rest masses.

The next useful concept is the relationship between binding energy per nucleon and nuclear stability. It explains why heavy nuclei can release energy through fission, while light nuclei can release energy through fusion.

For that broader explanation, continue with Energy Released in Fission and Fusion: HSC Physics Guide.

It is also worth keeping nuclear mass-energy conservation separate from mechanical energy conservation. The latter involves tracking kinetic and potential energy, as explained in Conservation of Mechanical Energy for HSC Physics.

Syllabus context: NESA’s Physics 11-12 Syllabus (2025) is scheduled for implementation from 2027, with the first HSC examination under the new syllabus in 2028. Check the syllabus version applying to your cohort when planning your study. See the official NESA implementation information.