How to Calculate Energy Released from a Nuclear Reaction’s Mass Change

A proton strikes a lithium-7 nucleus, and two helium-4 nuclei emerge. When you add up the rest masses before and after the reaction, the products are lighter by approximately 0.01862 atomic mass units (u).

That sounds like a tiny difference. But it corresponds to about 17.35 MeV of energy.

How do you calculate that energy, and how do you know whether the reaction releases energy or needs an input?

The method is straightforward:

1. Balance the nuclear equation using conservation of nucleon number and charge.
2. Calculate the total rest mass of the reactants.
3. Calculate the total rest mass of the products.
4. Subtract the final mass from the initial mass.
5. Convert the mass difference into energy using \(E=mc^2\).

The key equation is:

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

Here, \(Q\) is the energy released by the reaction, and \(c\) is the speed of light.

If the mass difference is measured in atomic mass units, you can use the much quicker conversion:

\[
\boxed{Q=\Delta m\times 931.5\ \mathrm{MeV}}
\]

where \(\Delta m\) is the initial mass minus the final mass, measured in u.

A positive value means energy is released. A negative value means energy must be supplied.

01Why physicists needed to measure the mass change

In 1932, John Cockcroft and Ernest Walton investigated whether accelerated protons could cause lithium nuclei to break apart.

There was an important question behind the experiment: if a nuclear reaction produces particles with more kinetic energy than the incoming particles, where does that extra energy come from?

Einstein’s mass-energy relationship suggested an answer. A decrease in the total rest mass of the system should correspond to an increase in other forms of energy.

But this needed experimental evidence.

How the Cockcroft-Walton experiment worked

Cockcroft and Walton constructed an accelerator that produced a beam of high-speed protons.

The apparatus worked in a particular sequence:

1. Hydrogen supplied protons, which were accelerated through a high-voltage tube.
2. The proton beam travelled towards a lithium target positioned at approximately 45 degrees to the beam.
3. The protons struck lithium nuclei, potentially producing alpha particles.
4. A thin mica window in the apparatus allowed alpha particles to escape.
5. A zinc sulphide screen was positioned opposite the window. When an alpha particle struck it, a tiny flash of light, called a scintillation, appeared.
6. An observer examined the screen through a microscope.

The mica window was important. It stopped scattered protons from reaching the detector while allowing the more penetrating alpha particles to pass through.

What should they observe if lithium nuclei were splitting into two alpha particles?

The zinc sulphide screen should produce flashes when alpha particles struck it. If two alpha particles were produced together, a suitable two-screen coincidence experiment should detect them as a pair.

The researchers initially investigated gamma radiation. They then changed their approach to detect alpha particles directly, using the mica window and scintillation screen.

The observations supported the production of alpha particles. Further experiments using two screens confirmed that the particles were emitted in pairs. Measurements of their ranges in air allowed their energies to be estimated.

Cockcroft and Walton measured approximately 8.6 MeV for each alpha particle, giving a total kinetic energy of about 17.2 MeV.

The reaction was interpreted as:

\[
{}^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{p}
\rightarrow
{}^{4}_{2}\mathrm{He}+{}^{4}_{2}\mathrm{He}
\]

This was important evidence that a decrease in nuclear rest mass could account for the energy released in a nuclear reaction.

There was also an interesting measurement problem. Their original mass estimates were not sufficiently accurate to produce a close numerical match. An early mass difference of 0.0154 u corresponded to approximately 14.3 MeV. A later, more accurate measurement of lithium-7 changed the estimated mass difference to 0.0180 u, corresponding to approximately 16.8 MeV.

The agreement with the measured energy improved considerably.

This illustrates an important experimental principle: when the mass difference is very small, even a modest error in the measured nuclear masses can produce a noticeable error in the calculated energy.

The historical details are described in Cockcroft’s Nobel lecture and the original 1932 Nature paper.

02Step 1: Balance the nuclear reaction

Before calculating any mass difference, make sure the nuclear equation is balanced.

Two quantities must be conserved:

• Nucleon number, \(A\): the total number of protons and neutrons.
• Nuclear charge, \(Z\): the total number of protons.

For the lithium reaction:

\[
{}^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{p}
\rightarrow
{}^{4}_{2}\mathrm{He}+{}^{4}_{2}\mathrm{He}
\]

Check the nucleon numbers:

\[
7+1=4+4=8
\]

Check the nuclear charges:

\[
3+1=2+2=4
\]

Both quantities are conserved, so the equation is balanced.

Remember that an alpha particle is a helium-4 nucleus containing two protons and two neutrons.

A neutron has \(A=1\) and \(Z=0\), while a proton has \(A=1\) and \(Z=1\).

If electrons or other leptons appear in a nuclear equation, their electric charges must also be included when checking charge conservation.

03Step 2: Calculate the mass difference

Once the equation is balanced, calculate the total rest mass on each side.

The mass difference is defined as:

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

In other words:

\[
\Delta m=
\text{total reactant mass}
–
\text{total product mass}
\]

This subtraction order matters.

If the products have less total rest mass than the reactants, \(\Delta m\) is positive. The difference in rest-mass energy is available to appear in other forms, such as kinetic energy or radiation.

04Step 3: Convert the mass difference into energy

Einstein’s mass-energy equation is:

\[
E=mc^2
\]

Here:

• \(E\) is energy, measured in joules (J).
• \(m\) is mass, measured in kilograms (kg).
• \(c\) is the speed of light, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

For nuclear reactions, using kilograms and joules is often inconvenient. Nuclear masses are commonly given in atomic mass units, so a conversion is useful.

The important conversion is:

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

Therefore:

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

One mega-electronvolt is:

\[
1\ \mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J}
\]

You can use this second conversion if a question requests the energy in joules.

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

These four terms are related, but they do not mean the same thing. Confusing them can lead to incorrect calculations.

Nuclear mass

The nuclear mass is the mass of the nucleus alone.

It includes:

• Protons.
• Neutrons.
• The effects of nuclear binding energy.

It does not include the electrons surrounding the nucleus.

Atomic mass

Atomic mass includes the nucleus and its surrounding electrons.

For a neutral atom:

\[
m_{\mathrm{atom}}\approx m_{\mathrm{nucleus}}+Zm_e
\]

Here, \(m_e\) is the electron mass and \(Z\) is the number of electrons in a neutral atom.

The exact relationship also accounts for the binding energy of the electrons, but this correction is generally negligible for HSC nuclear-energy calculations.

Atomic masses are often convenient because isotope mass tables commonly provide them.

Mass defect

The mass defect is the difference between the total mass of a nucleus’s separated nucleons and the actual mass of the bound nucleus.

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

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

Here:

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

The mass defect is related to the nuclear binding energy:

\[
B=\Delta m_{\mathrm{nucleus}}c^2
\]

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

Released energy

Released energy is different from the mass defect of an individual nucleus.

It is calculated by comparing the total rest masses of the entire reaction before and after it occurs:

\[
Q=(m_{\mathrm{reactants}}-m_{\mathrm{products}})c^2
\]

For reactions in which the total numbers of protons and neutrons are separately conserved, the same result can be calculated using total nuclear binding energies:

\[
Q=B_{\mathrm{products}}-B_{\mathrm{reactants}}
\]

This is because the total free-nucleon masses cancel.

For reactions involving beta decay, you must also account for the rest masses of the emitted leptons.

The important distinction is that the mass defect describes an individual bound nucleus, while the reaction mass difference compares the complete initial and final systems.

For a more detailed explanation of how nuclear mass defect relates to binding energy, see Mass defect: how missing mass becomes nuclear binding energy.

06Choosing between nuclear masses and atomic masses

Before substituting values into your equation, check which type of mass the question provides.

For ordinary nuclear reactions, total nuclear charge is conserved. If you represent every nuclear species using neutral atomic masses, the electron rest masses cancel because the total number of electrons is the same on both sides.

There is an important exception to remember: beta decay.

For beta-minus decay, a neutron converts into a proton, an electron and an electron antineutrino.

\[
{}^{1}_{0}\mathrm{n}
\rightarrow
{}^{1}_{1}\mathrm{p}
+
{}^{0}_{-1}\mathrm{e}
+
\bar{\nu}_e
\]

When calculating the energy released in beta-minus decay using neutral atomic masses, use:

\[
Q_{\beta^-}=
\left[
M_{\mathrm{parent}}-M_{\mathrm{daughter}}
\right]c^2
\]

Do not subtract the electron mass again. The neutral atomic mass difference already accounts for the electron rest-mass bookkeeping.

If you instead use bare nuclear masses, you must subtract the rest mass of the emitted electron.

This distinction becomes particularly important when the mass difference is small.

08What happens when the products have more mass?

So far, both worked examples have involved positive mass differences.

But what happens if the products are actually heavier than the reactants?

Consider the following reaction:

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

This reaction is balanced:

\[
A:14+4=17+1=18
\]

\[
Z:7+2=8+1=9
\]

Using the following neutral atomic masses:

Calculate the initial mass:

\[
\begin{aligned}
m_i&=14.003074+4.002603\\
&=18.005677\ \mathrm{u}
\end{aligned}
\]

Calculate the final mass:

\[
\begin{aligned}
m_f&=16.999132+1.007825\\
&=18.006957\ \mathrm{u}
\end{aligned}
\]

Find the mass difference:

\[
\begin{aligned}
\Delta m&=18.005677-18.006957\\
&=-0.001280\ \mathrm{u}
\end{aligned}
\]

Calculate the reaction energy:

\[
\begin{aligned}
Q&=-0.001280\times931.5\\
&=-1.19\ \mathrm{MeV}
\end{aligned}
\]

The negative sign is important.

The reaction requires an energy input of approximately 1.19 MeV in the centre-of-mass frame. It does not release energy.

For a projectile striking a stationary target, the required beam energy is greater than this value because momentum conservation also requires the reaction products to carry centre-of-mass motion.

A common mistake is to assume that every nuclear reaction releases energy because nuclear reactions involve mass changes.

That isn’t correct. The sign of the mass difference determines whether the reaction releases energy or requires an input.

09Questions and solutions

Try these questions without looking at the solutions first. They progress from a standard mass-difference calculation to problems involving binding energy, momentum conservation and electron bookkeeping.

Question 1

Sodium-23 captures a neutron to form sodium-24 and a gamma photon.

\[
{}^{23}_{11}\mathrm{Na}
+
{}^{1}_{0}\mathrm{n}
\rightarrow
{}^{24}_{11}\mathrm{Na}
+
\gamma
\]

Using the following rounded neutral atomic masses (illustrative data), calculate the energy released per reaction in MeV.

Solution 1

The reaction releases approximately 6.96 MeV per capture.

Step 1: Confirm conservation of nucleon number and charge.

Nucleon number:

\[
23+1=24
\]

Nuclear charge:

\[
11+0=11
\]

The equation is balanced.

Step 2: Calculate the total initial mass.

\[
\begin{aligned}
m_i&=22.989770+1.008665\\
&=23.998435\ \mathrm{u}
\end{aligned}
\]

Step 3: Calculate the total final mass.

\[
m_f=23.990963\ \mathrm{u}
\]

Step 4: Calculate the mass difference.

\[
\begin{aligned}
\Delta m&=23.998435-23.990963\\
&=0.007472\ \mathrm{u}
\end{aligned}
\]

Step 5: Convert into energy.

\[
\begin{aligned}
Q&=0.007472\times931.5\\
&=6.96\ \mathrm{MeV}
\end{aligned}
\]

The positive value means energy is released.

The energy is carried by the gamma radiation and the recoil of the sodium-24 nucleus. The gamma photon does not necessarily carry the entire Q value because some energy goes into recoil.

Question 2

An alpha particle strikes an aluminium-27 nucleus. The reaction produces phosphorus-30 and one other particle.

\[
{}^{27}_{13}\mathrm{Al}
+
{}^{4}_{2}\mathrm{He}
\rightarrow
{}^{30}_{15}\mathrm{P}
+
X
\]

(a) Determine the identity of particle \(X\) using conservation of nucleon number and charge.

(b) Using the following rounded neutral atomic masses (illustrative data), calculate the Q value of the reaction.

(c) State whether the reaction releases or absorbs energy.

Solution 2

The missing particle is a neutron, and the reaction absorbs approximately 2.64 MeV.

Step 1: Use conservation of nucleon number to identify the missing particle.

The initial nucleon number is:

\[
A=27+4=31
\]

Phosphorus-30 accounts for 30 nucleons.

Therefore, the missing particle must have:

\[
A=31-30=1
\]

Step 2: Use conservation of charge.

The initial nuclear charge is:

\[
Z=13+2=15
\]

Phosphorus-30 already has a nuclear charge of 15.

Therefore, the missing particle must have:

\[
Z=0
\]

A particle with \(A=1\) and \(Z=0\) is a neutron.

Step 3: Calculate the total initial mass.

\[
\begin{aligned}
m_i&=26.981539+4.002603\\
&=30.984142\ \mathrm{u}
\end{aligned}
\]

Step 4: Calculate the total final mass.

\[
\begin{aligned}
m_f&=29.978314+1.008665\\
&=30.986979\ \mathrm{u}
\end{aligned}
\]

Step 5: Calculate the mass difference.

\[
\begin{aligned}
\Delta m&=30.984142-30.986979\\
&=-0.002837\ \mathrm{u}
\end{aligned}
\]

Step 6: Calculate the Q value.

\[
\begin{aligned}
Q&=-0.002837\times931.5\\
&=-2.64\ \mathrm{MeV}
\end{aligned}
\]

The negative Q value means the products have greater total rest mass than the reactants.

At least 2.64 MeV of energy must be supplied in the centre-of-mass frame for the reaction to occur. A stationary-target experiment requires a higher incident beam energy because momentum must also be conserved.

Question 3

Lithium-6 reacts with deuterium to produce two helium-4 nuclei.

\[
{}^{6}_{3}\mathrm{Li}
+
{}^{2}_{1}\mathrm{H}
\rightarrow
2{}^{4}_{2}\mathrm{He}
\]

The following total nuclear binding energies are provided (rounded illustrative data):

(a) Calculate the total binding energy of the reactants.

(b) Calculate the total binding energy of the products.

(c) Determine the energy released by the reaction.

(d) Explain why comparing the binding energy of just one reactant with the binding energy of one product would give an incorrect result.

Solution 3

The reaction releases approximately 22.37 MeV.

Step 1: Calculate the total reactant binding energy.

There are two reactants: lithium-6 and deuterium.

\[
\begin{aligned}
B_i&=31.995+2.225\\
&=34.220\ \mathrm{MeV}
\end{aligned}
\]

Step 2: Calculate the total product binding energy.

There are two helium-4 nuclei:

\[
\begin{aligned}
B_f&=2(28.296)\\
&=56.592\ \mathrm{MeV}
\end{aligned}
\]

Step 3: Calculate the energy released.

For this reaction, the total numbers of protons and neutrons are separately conserved. Therefore, the reaction Q value equals the increase in total binding energy:

\[
Q=B_f-B_i
\]

Substitute:

\[
\begin{aligned}
Q&=56.592-34.220\\
&=22.372\ \mathrm{MeV}
\end{aligned}
\]

The reaction releases approximately:

\[
\boxed{Q=22.37\ \mathrm{MeV}}
\]

Step 4: Explain why total binding energy matters.

The binding energy describes how strongly each nucleus holds its nucleons together.

It is tempting to compare the binding energy of lithium-6 with that of one helium-4 nucleus. However, there are two helium-4 nuclei in the products, and both contribute to the final total binding energy.

The correct comparison must include every nucleus on both sides.

The products have a greater total binding energy, meaning they have a lower combined rest mass than the reactants. The difference is released as energy.

Question 4

A hypothetical nuclear reaction occurs in its centre-of-mass frame.

The reactants have negligible initial relative kinetic energy, and the reaction releases 4.80 MeV. The reaction produces two particles, C and D, with rest masses of 4 u and 12 u, respectively.

Assume no gamma photons, neutrinos or other particles are produced.

(a) Calculate the total kinetic energy of the products.

(b) Determine the ratio of the kinetic energies of C and D.

(c) Calculate the kinetic energy of each particle.

A student predicts that each particle receives 2.40 MeV because there are two products. Explain why this prediction is incorrect.

Solution 4

Particle C receives 3.60 MeV, while particle D receives 1.20 MeV.

The energy is not divided equally because the particles must also obey conservation of momentum.

Step 1: Determine the total kinetic energy available.

The initial relative kinetic energy is negligible, and no other energy carriers are produced.

Therefore, the released energy becomes the total kinetic energy of the products:

\[
K_C+K_D=4.80\ \mathrm{MeV}
\]

Step 2: Apply conservation of momentum.

The initial total momentum in the centre-of-mass frame is zero.

Therefore, the two products must have equal momentum magnitudes in opposite directions:

\[
p_C=p_D
\]

Step 3: Relate kinetic energy to momentum.

At these energies, the products are moving sufficiently slowly compared with light that the non-relativistic kinetic energy equation is appropriate:

\[
K=\frac{p^2}{2m}
\]

Since both particles have equal momentum magnitudes:

\[
\frac{K_C}{K_D}=\frac{m_D}{m_C}
\]

Substitute the masses:

\[
\begin{aligned}
\frac{K_C}{K_D}
&=\frac{12}{4}\\
&=3
\end{aligned}
\]

Therefore:

\[
K_C=3K_D
\]

Step 4: Calculate each kinetic energy.

Substitute into the total kinetic energy equation:

\[
3K_D+K_D=4.80
\]

\[
4K_D=4.80
\]

\[
K_D=1.20\ \mathrm{MeV}
\]

Therefore:

\[
K_C=3(1.20)=3.60\ \mathrm{MeV}
\]

The final results are:

\[
\boxed{K_C=3.60\ \mathrm{MeV}}
\]

\[
\boxed{K_D=1.20\ \mathrm{MeV}}
\]

The tempting equal split ignores momentum conservation.

Both particles have equal and opposite momentum, but the lighter particle must move faster. Since kinetic energy depends on both mass and velocity, the lighter particle receives more kinetic energy.

Question 5

A hypothetical beta-minus decay occurs:

\[
{}^{14}_{6}\mathrm{C}
\rightarrow
{}^{14}_{7}\mathrm{N}^{*}
+
{}^{0}_{-1}\mathrm{e}
+
\bar{\nu}_e
\]

The nitrogen-14 daughter is initially produced in an excited state, 0.100 MeV above its ground state. It subsequently emits a gamma photon of 0.100 MeV to reach the ground state.

The following neutral atomic masses are provided:

Additional information:

• Electron mass: \(m_e=0.000548580\ \mathrm{u}\).
• \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).
• Ignore electron binding energy corrections and neutrino rest mass.

Two students attempt to calculate the energy released.

• Student A uses the neutral atomic masses and subtracts the electron mass again because a beta electron appears in the products.
• Student B calculates the nuclear masses but forgets to include the emitted electron’s rest mass.

Both students obtain different answers.

(a) Calculate the total Q value using the neutral atomic masses.

(b) Explain why Student A’s method is incorrect.

(c) Show how the calculation would be performed using nuclear masses instead.

(d) Determine whether production of the excited nitrogen-14 state is energetically possible.

(e) Calculate the maximum possible kinetic energy of the emitted beta electron, assuming the antineutrino and nuclear recoil carry negligible energy.

Solution 5

The beta-minus decay is energetically allowed. The total ground-state Q value is approximately 0.1565 MeV, but only 0.0565 MeV is available to the beta electron, antineutrino and recoil when the daughter is produced in the specified excited state.

The crucial distinction is that neutral atomic masses already account for the electron rest-mass bookkeeping.

Step 1: Check conservation of nucleon number and charge.

Nucleon number:

\[
14=14+0+0
\]

Charge:

\[
6=7-1+0
\]

Both quantities are conserved.

Step 2: Calculate the mass difference using atomic masses.

\[
\begin{aligned}
\Delta m&=m_{\mathrm{C-14}}-m_{\mathrm{N-14}}\\
&=14.003241988-14.003074004\\
&=0.000167984\ \mathrm{u}
\end{aligned}
\]

Step 3: Calculate the total Q value.

\[
\begin{aligned}
Q&=0.000167984\times931.5\\
&=0.1565\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
\boxed{Q=0.1565\ \mathrm{MeV}}
\]

This is the total energy available for the decay to the daughter ground state.

Step 4: Explain why Student A is incorrect.

Student A subtracts the electron mass from the atomic mass difference:

\[
\Delta m_{\mathrm{wrong}}
=
0.000167984-0.000548580
\]

This gives:

\[
\Delta m_{\mathrm{wrong}}
=-0.000380596\ \mathrm{u}
\]

The resulting Q value would be approximately:

\[
Q_{\mathrm{wrong}}=-0.3545\ \mathrm{MeV}
\]

This would incorrectly suggest that the decay requires energy input.

The problem is double-counting the electron mass.

When neutral atomic masses are used for beta-minus decay, the extra electron in the daughter atom’s mass already accounts for the rest-mass bookkeeping associated with the emitted beta electron.

You must not subtract its mass again.

Step 5: Show the correct nuclear-mass calculation.

If using nuclear masses, first remove the electron masses from the neutral atomic masses.

For carbon-14:

\[
\begin{aligned}
m_{\mathrm{C,nucleus}}
&=14.003241988-6(0.000548580)\\
&=13.999950508\ \mathrm{u}
\end{aligned}
\]

For nitrogen-14:

\[
\begin{aligned}
m_{\mathrm{N,nucleus}}
&=14.003074004-7(0.000548580)\\
&=13.999233944\ \mathrm{u}
\end{aligned}
\]

Now include the emitted electron’s rest mass:

\[
Q=
(m_{\mathrm{C,nucleus}}
-m_{\mathrm{N,nucleus}}
-m_e)c^2
\]

Substitute:

\[
\begin{aligned}
\Delta m
&=13.999950508\\
&\quad-13.999233944\\
&\quad-0.000548580\\
&=0.000167984\ \mathrm{u}
\end{aligned}
\]

Therefore:

\[
Q=0.000167984\times931.5
\]

\[
\boxed{Q=0.1565\ \mathrm{MeV}}
\]

This agrees with the atomic-mass calculation.

Student B’s mistake is forgetting to subtract the emitted electron mass when using bare nuclear masses. That would overestimate the available energy by approximately 0.511 MeV.

Step 6: Account for the excited daughter state.

The nitrogen-14 nucleus is produced 0.100 MeV above its ground state.

That excitation energy must come from the total Q value.

Therefore, the energy available to the beta decay products is:

\[
\begin{aligned}
Q_{\mathrm{excited}}
&=Q-E_{\mathrm{excitation}}\\
&=0.1565-0.100\\
&=0.0565\ \mathrm{MeV}
\end{aligned}
\]

Since this value is positive, the decay to the excited state is energetically possible.

Step 7: Calculate the maximum beta-electron kinetic energy.

The available 0.0565 MeV is shared between:

• The beta electron’s kinetic energy.
• The antineutrino’s energy.
• The recoil kinetic energy of the daughter nucleus.

The beta electron therefore cannot receive more than the available 0.0565 MeV.

If the antineutrino and recoil carry negligible energy, the maximum beta-electron kinetic energy is approximately:

\[
\boxed{K_{\mathrm{e,max}}=0.0565\ \mathrm{MeV}}
\]

Or:

\[
\boxed{K_{\mathrm{e,max}}=56.5\ \mathrm{keV}}
\]

The beta electron will generally receive less because the antineutrino and daughter recoil also carry energy.

The excited nitrogen-14 nucleus subsequently releases its 0.100 MeV excitation energy as a gamma photon.

This is the important distinction: the original mass difference determines the total energy available, but the energy does not necessarily all appear as kinetic energy of one emitted particle.

10Where to go next

Once you can calculate the Q value of a nuclear reaction, the next step is understanding where that energy goes.

It might become kinetic energy of reaction products, gamma radiation, or other emitted particles. Momentum conservation and the reaction products’ masses help determine how that energy is distributed.

This is particularly useful when studying fission and fusion, where the difference between the total binding energies of the reactants and products determines whether energy is released.

For the next application, see Energy Released in Fission and Fusion: HSC Physics Guide.