Why Gamma Emission Does Not Change an Element’s Identity

An excited nucleus containing 27 protons and 33 neutrons emits a gamma photon. It loses energy, and its rest mass decreases. Yet after emission, it is still cobalt-60. It has not become cobalt-59 or nickel-60.

How can a nucleus lose mass without changing its identity?

The answer is that gamma emission changes the energy state of a nucleus, not its number of protons or neutrons. An element is identified by its proton number, \(Z\), while an isotope is identified by its proton and neutron numbers. Neither changes during pure gamma emission.

The important distinction is between the number of nucleons in a nucleus and the actual mass of that nucleus. They are related, but they are not the same quantity.

01The nuclear bookkeeping: what makes an element the same element?

Start with the notation used to represent a nucleus:

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

Each part tells you something different:

• \(X\) is the chemical symbol of the element.
• \(Z\) is the atomic number, equal to the number of protons.
• \(A\) is the mass number, equal to the total number of protons and neutrons.
• \(N=A-Z\) is the number of neutrons.

The atomic number determines the element. For example, every nucleus containing 27 protons is cobalt, regardless of how many neutrons it contains.

The mass number helps distinguish isotopes. Cobalt-59 and cobalt-60 are both cobalt because they each have 27 protons, but they contain different numbers of neutrons.

Now consider an excited nucleus undergoing gamma emission.

Its nuclear equation is:

\[
{}^{A}_{Z}X^*
\rightarrow
{}^{A}_{Z}X+\gamma
\]

The asterisk indicates that the original nucleus is in an excited energy state. It does not represent a different isotope.

Notice what happens to the two numbers:

• The mass number remains \(A\).
• The atomic number remains \(Z\).

The gamma photon has zero electric charge and zero nucleon number. It carries energy, but it does not remove a proton or neutron.

We can therefore check both conservation laws:

Conservation of nucleon number:

\[
A=A+0
\]

Conservation of nuclear charge:

\[
Z=Z+0
\]

Both are satisfied.

Why this differs from alpha and beta decay

It helps to compare gamma emission with two other common types of radioactive decay.

In alpha decay, the nucleus actually loses four nucleons. Its proton number also decreases by two, so it becomes a different element.

In beta-minus decay, a neutron transforms into a proton, producing an electron and an electron antineutrino:

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

The total number of nucleons stays constant, but the proton number increases by one. The element therefore changes.

Gamma emission does neither. No nucleons are removed, and no neutron changes into a proton.

This gives you a useful rule for nuclear equations:

If only gamma radiation is emitted, the element and isotope remain unchanged.

However, there is a complication. If the nucleus loses energy, shouldn’t its mass decrease?

Yes. This is where mass number and nuclear mass need to be separated.

02Why the nucleus loses mass without losing nucleons

Imagine a ball sitting on a high shelf. It has gravitational potential energy because of its position.

If the ball falls, it releases energy, but it doesn’t lose a piece of itself. It is still the same ball.

An excited nucleus behaves somewhat similarly. It contains additional internal energy compared with its lower-energy state. When it releases that energy, its composition stays the same.

The analogy has a limit: nuclear energy comes from quantum energy states and interactions between nucleons, not from the nucleus being physically positioned higher in a gravitational field.

Nuclear energy and mass are connected

According to Einstein’s mass-energy equivalence:

\[
E=mc^2
\]

where:

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

This equation tells us that energy contributes to the rest mass of a system.

An excited nucleus has more internal energy than the same nucleus in its ground state. Consequently, it also has a greater rest mass.

Let:

• \(m_*\) be the nuclear rest mass in the excited state.
• \(m_g\) be the nuclear rest mass in the ground state.
• \(E_{\mathrm{exc}}\) be the excitation energy above the ground state.

Then:

\[
m_*=m_g+\frac{E_{\mathrm{exc}}}{c^2}
\]

When the nucleus emits a gamma photon, it moves to a lower energy state.

Its rest mass decreases:

\[
m_*>m_g
\]

The energy difference is released as electromagnetic radiation.

A gamma photon has zero rest mass, but it does carry energy and momentum.

Its energy is:

\[
E_\gamma=hf
\]

where \(h\) is Planck’s constant and \(f\) is the photon frequency.

For an idealised calculation where nuclear recoil is negligible, the energy released is approximately:

\[
E_\gamma=(m_*-m_g)c^2
\]

This means the nucleus loses a tiny amount of rest mass when it emits the photon.

There is no contradiction:

• The mass number stays constant because no nucleons are lost.
• The nuclear rest mass decreases because the nucleus loses energy.
• The element stays the same because its proton number is unchanged.

One further detail matters for more precise calculations. A photon carries momentum, so an isolated nucleus initially at rest must recoil in the opposite direction when it emits a photon.

Therefore, the exact energy balance is:

\[
(m_*-m_g)c^2=E_\gamma+K_{\mathrm{recoil}}
\]

Here, \(K_{\mathrm{recoil}}\) is the kinetic energy of the recoiling nucleus.

For most HSC calculations, this recoil energy is negligible unless the question specifically asks you to consider it.

03Nuclear mass, atomic mass, mass defect and binding energy

These four terms are closely related, but they describe different things. Understanding the differences helps explain why gamma emission changes mass without changing an element.

Let’s connect these ideas to gamma emission.

1. Nuclear mass

The nuclear mass is the actual rest mass of the nucleus.

It is not simply the number of nucleons multiplied by the mass of one nucleon.

For a nucleus containing \(Z\) protons and \(N\) neutrons, the combined mass of the separate particles would be:

\[
m_{\mathrm{separate}}=Zm_p+Nm_n
\]

where:

• \(m_p\) is the mass of one free proton.
• \(m_n\) is the mass of one free neutron.

The actual bound nucleus has a smaller rest mass than this combined mass.

That difference is related to the energy released when the nucleus formed.

2. Atomic mass

Atomic mass includes the nucleus and its surrounding electrons.

For a neutral atom containing \(Z\) protons, there are also \(Z\) electrons.

A useful approximation is:

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

where \(m_e\) is the electron mass.

The exact relationship also includes the binding energy of the electrons.

Why does this distinction matter for gamma emission?

Suppose you compare the atomic mass of an excited neutral atom with the atomic mass of the same atom after gamma emission.

Both atoms have the same proton number, so they also have the same number of electrons.

Their electron rest-mass contributions cancel when you subtract the two atomic masses.

Therefore, provided the electronic state is unchanged, the difference between the atomic masses reflects the nuclear energy change to HSC calculation precision.

You do not need to remove the electron masses when comparing two neutral states of the same isotope.

3. Mass defect

The mass defect is the difference between the total mass of the separate nucleons and the actual nuclear mass.

It is calculated using:

\[
\Delta m_{\mathrm{defect}}
=
Zm_p+Nm_n-m_{\mathrm{nucleus}}
\]

The mass defect is positive for a bound nucleus.

An excited nucleus has a greater rest mass than the same nucleus in its ground state. Therefore, its mass defect is smaller.

When gamma emission occurs, the nuclear mass decreases and the mass defect increases.

4. Binding energy

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

It is related to mass defect by:

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

where \(B\) is the total nuclear binding energy.

Now consider the relationship between an excited nucleus and its ground state.

Both contain exactly the same protons and neutrons. Therefore, the combined mass of their separate nucleons is identical.

However, the excited nucleus has a greater actual nuclear mass. Its binding energy is consequently smaller.

When it emits a gamma photon and reaches the ground state, its binding energy increases.

Neglecting recoil:

\[
B_g-B_*\approx E_\gamma
\]

This is an important result.

Gamma emission increases the binding energy of the nucleus without changing the number of nucleons.

The nucleus becomes more tightly bound, but it remains the same isotope.

5. Released energy

Released energy is not another name for mass defect or total binding energy.

It describes the energy transferred out of a system during a process.

For a nuclear transition, we can define the released energy as:

\[
Q=(m_i-m_f)c^2
\]

where:

• \(m_i\) is the initial nuclear rest mass.
• \(m_f\) is the final nuclear rest mass.
• \(Q\) is positive when energy is released.

For gamma emission, this energy becomes the gamma photon energy and, in an isolated system, a small amount of recoil kinetic energy.

This distinction is particularly useful in exam questions:

• Mass defect describes the difference between separate nucleons and a bound nucleus.
• Binding energy describes how much energy is needed to separate those nucleons.
• Released energy describes the energy transferred during a particular process.
• Nuclear mass describes the actual rest mass of the nucleus.

If you need to calculate an absolute binding energy from nuclear or atomic mass data, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

04How experiments measure the energy released by gamma emission

There is an experimental question behind this explanation:

How can we establish that gamma emission involves specific changes in nuclear energy rather than the nucleus randomly releasing any amount of energy?

A standard modern method is gamma-ray spectroscopy.

Building the experiment

Imagine a radioactive source containing excited nuclei.

The source is placed inside lead shielding, with a narrow opening that allows gamma photons to travel towards a detector.

A high-purity germanium (HPGe) detector is positioned beyond this opening.

The apparatus works in the following sequence:

1. Excited nuclei in the radioactive source undergo nuclear transitions and emit gamma photons.
2. The photons travel through the opening in the lead shielding towards the detector.
3. A photon interacts with the germanium crystal, transferring energy to charged particles in the detector.
4. This produces an electrical signal whose size is related to the energy deposited.
5. A preamplifier and signal-processing electronics prepare the signal for measurement.
6. A multichannel analyser sorts the signals into energy ranges and builds a gamma-ray spectrum.

The spectrum is a graph showing the number of detected events at different photon energies.

What should the experiment show?

If nuclear energy levels are discrete, we should expect gamma photons with particular energies corresponding to transitions between those levels.

We would therefore expect distinct peaks in the measured spectrum rather than every possible photon energy appearing equally.

This is what gamma spectroscopy reveals: peaks at particular energies.

A nucleus can also undergo a cascade of transitions, emitting multiple gamma photons as it moves through several energy levels.

The observed photon energies provide information about the differences between those levels.

Not every detected event produces a full-energy peak. Some photons transfer only part of their energy to the detector, producing a background known as the Compton continuum.

This is why detector resolution and calibration matter.

Why a simple counter is not enough

An ordinary Geiger-Müller counter can detect radiation and measure count rates.

However, it cannot resolve individual gamma photon energies with the precision needed to map nuclear energy levels.

It tells us that radiation has arrived, but not accurately how much energy each photon carried.

An energy-resolving detector such as an HPGe detector addresses this limitation.

Calibration using known gamma-ray energies allows the measured peak positions to be converted into photon energies.

The result supports the model of discrete nuclear energy states.

There is an important limitation: a gamma spectrum measures photon energies, not the number of protons and neutrons in the emitting nucleus. Establishing the nuclide’s identity also requires knowledge of the parent and daughter nuclei and, where necessary, independent isotope identification.

The spectroscopy provides evidence about the energy transition. Nuclear bookkeeping establishes why pure gamma emission leaves \(A\) and \(Z\) unchanged.

For a broader explanation of gamma radiation, including its penetration and ionisation properties, see Gamma Decay Explained: Emission, Penetration and Ionisation.

05Worked examples

06Questions and solutions

Question 1

A bromine-81 nucleus in an excited state undergoes gamma emission.

\[
{}^{81}_{35}\mathrm{Br}^*
\rightarrow
{}^{81}_{35}\mathrm{Br}+\gamma
\]

Determine:

(a) The number of protons before and after emission.

(b) The number of neutrons before and after emission.

(c) The mass number and identity of the resulting nucleus.

Solution 1

The resulting nucleus is still bromine-81, containing 35 protons and 46 neutrons.

Gamma emission does not remove nucleons or change the proton number.

Part (a): Protons

The atomic number is:

\[
Z=35
\]

Therefore, there are 35 protons before and after emission.

Part (b): Neutrons

Use:

\[
N=A-Z
\]

Substitute:

\[
N=81-35=46
\]

There are 46 neutrons before and after emission.

Part (c): Mass number and identity

The mass number remains:

\[
A=81
\]

The atomic number remains:

\[
Z=35
\]

Therefore, the resulting nucleus is:

\[
{}^{81}_{35}\mathrm{Br}
\]

The important point is that gamma emission changes the nuclear energy state, not the nuclear composition.

Question 2

A hypothetical nucleus has three energy levels:

• Ground state: \(0.00\ \mathrm{MeV}\)
• Intermediate state: \(0.45\ \mathrm{MeV}\)
• Upper excited state: \(1.20\ \mathrm{MeV}\)

The nucleus undergoes a two-stage gamma transition, moving first from the upper state to the intermediate state, then from the intermediate state to the ground state.

Determine:

(a) The energy of the first gamma photon.

(b) The energy of the second gamma photon.

(c) The total energy released.

(d) The equivalent decrease in nuclear rest mass, in atomic mass units.

(e) Whether the isotope changes after the first transition.

Assume nuclear recoil is negligible.

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

Solution 2

The two gamma photons have energies of \(0.75\ \mathrm{MeV}\) and \(0.45\ \mathrm{MeV}\), giving a total released energy of \(1.20\ \mathrm{MeV}\). The isotope remains unchanged after both transitions.

Part (a): First gamma photon

The first transition is from \(1.20\ \mathrm{MeV}\) to \(0.45\ \mathrm{MeV}\).

The photon receives the difference between these energy levels:

\[
\begin{aligned}
E_{\gamma 1}
&=1.20-0.45\\
&=0.75\ \mathrm{MeV}
\end{aligned}
\]

Part (b): Second gamma photon

The second transition is from \(0.45\ \mathrm{MeV}\) to the ground state.

\[
\begin{aligned}
E_{\gamma 2}
&=0.45-0.00\\
&=0.45\ \mathrm{MeV}
\end{aligned}
\]

Part (c): Total energy released

Add the energies of both photons:

\[
\begin{aligned}
E_{\mathrm{total}}
&=0.75+0.45\\
&=1.20\ \mathrm{MeV}
\end{aligned}
\]

This matches the original energy difference between the upper state and the ground state.

Part (d): Equivalent decrease in nuclear rest mass

Use:

\[
\Delta m=\frac{E_{\mathrm{total}}}{931.5}
\]

Substitute:

\[
\begin{aligned}
\Delta m
&=\frac{1.20}{931.5}\\
&=1.29\times10^{-3}\ \mathrm{u}
\end{aligned}
\]

The nucleus loses approximately \(1.29\times10^{-3}\ \mathrm{u}\) of rest mass.

Part (e): Does the isotope change after the first transition?

No. The nucleus remains the same isotope after the first photon is emitted.

It simply moves from the upper energy state to the intermediate energy state.

It still contains exactly the same number of protons and neutrons.

A common mistake is to assume that emitting the first photon means the nucleus has completed its decay. It has only completed one energy transition. It still has additional excitation energy to release.

Question 3

A uranium-238 nucleus undergoes alpha decay, leaving its daughter nucleus in an excited state.

The excited daughter then emits a gamma photon with an energy of \(0.090\ \mathrm{MeV}\).

Determine:

(a) The daughter nucleus immediately after alpha decay.

(b) The final nucleus after gamma emission.

(c) The number of protons and neutrons in the final nucleus.

(d) The equivalent decrease in nuclear rest mass due to gamma emission.

(e) Explain which process changes the element’s identity.

Assume recoil is negligible and use \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).

Solution 3

The final nucleus is thorium-234, containing 90 protons and 144 neutrons. Alpha decay changes the element from uranium to thorium, while the subsequent gamma emission changes only the energy state.

Part (a): Daughter after alpha decay

An alpha particle contains two protons and two neutrons.

Therefore, alpha decay reduces the mass number by four and the atomic number by two.

\[
{}^{238}_{92}\mathrm{U}
\rightarrow
{}^{234}_{90}\mathrm{Th}^*
+
{}^{4}_{2}\mathrm{He}
\]

The daughter is excited thorium-234.

Part (b): Final nucleus after gamma emission

The excited thorium nucleus releases its excess energy:

\[
{}^{234}_{90}\mathrm{Th}^*
\rightarrow
{}^{234}_{90}\mathrm{Th}+\gamma
\]

Neither its mass number nor its atomic number changes.

The final nucleus is thorium-234.

Part (c): Number of protons and neutrons

The atomic number gives the proton count:

\[
Z=90
\]

The neutron count is:

\[
\begin{aligned}
N&=A-Z\\
&=234-90\\
&=144
\end{aligned}
\]

The final nucleus contains 90 protons and 144 neutrons.

Part (d): Equivalent mass decrease due to gamma emission

Use:

\[
\Delta m=\frac{E_\gamma}{931.5}
\]

Substitute:

\[
\begin{aligned}
\Delta m
&=\frac{0.090}{931.5}\\
&=9.66\times10^{-5}\ \mathrm{u}
\end{aligned}
\]

The excited thorium nucleus loses approximately \(9.66\times10^{-5}\ \mathrm{u}\) of rest mass.

This does not change its mass number of 234.

Part (e): Which process changes the element?

Alpha decay changes the element because it removes two protons.

The atomic number decreases from 92 to 90, changing uranium into thorium.

Gamma emission does not change the element because it removes no protons.

This is an important distinction when a nuclear decay occurs in multiple stages. You must examine each process separately rather than treating the entire sequence as one type of decay.

Question 4

A hypothetical neutral isotope has a mass number of 84 and an atomic number of 36.

Its measured atomic masses are:

A student argues that the difference in atomic mass must be caused by the atom losing electrons.

The electron mass is \(5.49\times10^{-4}\ \mathrm{u}\).

(a) Calculate the energy released during the gamma transition.

(b) Explain why the atom does not need to lose any electrons.

(c) Determine the approximate change in total nuclear binding energy.

(d) Explain why subtracting the mass of 36 electrons from only one of the supplied atomic masses would be incorrect.

Assume the electronic configuration is unchanged and neglect nuclear recoil and changes in electronic binding energy.

Solution 4

The transition releases approximately \(0.335\ \mathrm{MeV}\), and the atom does not lose electrons. Its total nuclear binding energy increases by approximately \(0.335\ \mathrm{MeV}\).

Part (a): Calculate the released energy

First calculate the atomic mass difference:

\[
\begin{aligned}
\Delta m
&=m_i-m_f\\
&=83.911100-83.910740\\
&=0.000360\ \mathrm{u}
\end{aligned}
\]

Convert this mass difference into energy:

\[
Q=\Delta m\times931.5
\]

\[
\begin{aligned}
Q&=0.000360\times931.5\\
&=0.33534\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
Q\approx0.335\ \mathrm{MeV}
\]

Part (b): Does the atom lose electrons?

No.

The atomic number remains 36, so the nucleus still contains 36 protons.

Both the initial and final atoms are neutral. Each therefore contains 36 electrons.

The energy release comes from the nuclear transition, not from the removal of electrons.

Part (c): Change in nuclear binding energy

The initial and final states contain the same number of protons and neutrons.

Their separate-nucleon reference mass is therefore unchanged.

The decrease in nuclear mass corresponds to an increase in binding energy:

\[
\Delta B=B_f-B_i
\]

Neglecting recoil:

\[
\Delta B\approx0.335\ \mathrm{MeV}
\]

The ground-state nucleus is more tightly bound than the excited nucleus.

Part (d): Why should you not subtract 36 electron masses from only one atomic mass?

Both supplied values are atomic masses.

Each already includes the mass contributions of 36 electrons.

When the two atomic masses are subtracted, these electron contributions cancel.

Subtracting the electron masses from only one value would introduce an artificial mass difference that does not represent the nuclear transition.

The key is to use consistent mass types. Comparing two neutral atomic masses of the same isotope is valid here because their electron contributions are the same.

Question 5

An isolated excited nucleus of cobalt-60 is initially at rest:

\[
{}^{60}_{27}\mathrm{Co}^*
\]

It emits a gamma photon with an energy of \(0.800\ \mathrm{MeV}\).

After emission, the nucleus recoils in the opposite direction to the photon.

For the recoil calculation, use:

• Final nuclear mass: \(M_f=60\ \mathrm{u}\).
• Gamma photon energy: \(E_\gamma=0.800\ \mathrm{MeV}\).
• \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).

Assume non-relativistic nuclear recoil.

A student makes the following claims:

1. The mass number must decrease because the nucleus loses mass.
2. The decrease in nuclear rest-mass energy must be exactly \(0.800\ \mathrm{MeV}\).
3. The nuclear binding energy cannot change because the number of protons and neutrons stays constant.
4. Since the gamma photon has zero rest mass, it cannot account for any decrease in the nucleus’s rest mass.

Evaluate all four claims.

Calculate:

(a) The recoil kinetic energy of the nucleus.

(b) The decrease in nuclear rest-mass energy.

(c) The corresponding decrease in nuclear rest mass in atomic mass units.

(d) The change in nuclear binding energy.

Solution 5

All four claims are incorrect or incomplete. The nucleus remains cobalt-60, its rest-mass energy decreases by slightly more than \(0.800\ \mathrm{MeV}\), and its binding energy increases.

The subtle point is that the emitted photon carries momentum as well as energy. The nucleus must recoil, so not all the released energy becomes photon energy.

Part (a): Calculate the recoil kinetic energy

The nucleus is initially at rest, so its initial momentum is zero.

Momentum must be conserved.

The photon travels in one direction, so the nucleus recoils in the opposite direction with equal momentum magnitude.

For a photon:

\[
p_\gamma=\frac{E_\gamma}{c}
\]

Therefore:

\[
p_R=p_\gamma=\frac{E_\gamma}{c}
\]

Using the non-relativistic kinetic energy equation:

\[
K_R=\frac{p_R^2}{2M_f}
\]

Substitute the photon momentum:

\[
K_R=\frac{E_\gamma^2}{2M_fc^2}
\]

First calculate the final nuclear rest energy:

\[
\begin{aligned}
M_fc^2&=60\times931.5\\
&=55\,890\ \mathrm{MeV}
\end{aligned}
\]

Now substitute:

\[
\begin{aligned}
K_R
&=\frac{(0.800)^2}{2(55\,890)}\\
&=5.72\times10^{-6}\ \mathrm{MeV}
\end{aligned}
\]

Convert to electronvolts:

\[
K_R=5.72\ \mathrm{eV}
\]

The recoiling nucleus receives approximately \(5.72\ \mathrm{eV}\) of kinetic energy.

Part (b): Calculate the decrease in nuclear rest-mass energy

The initial nuclear rest-mass energy is converted into two forms:

• Energy carried by the gamma photon.
• Kinetic energy of the recoiling nucleus.

Therefore:

\[
\Delta E_{\mathrm{rest}}=E_\gamma+K_R
\]

Substitute:

\[
\begin{aligned}
\Delta E_{\mathrm{rest}}
&=0.800000+0.00000572\\
&=0.80000572\ \mathrm{MeV}
\end{aligned}
\]

The decrease in nuclear rest-mass energy is approximately:

\[
\Delta E_{\mathrm{rest}}=0.80000572\ \mathrm{MeV}
\]

The recoil contribution is tiny compared with the gamma photon energy, which explains why HSC calculations generally neglect it.

Part (c): Calculate the decrease in nuclear rest mass

Use mass-energy equivalence:

\[
\Delta m=\frac{\Delta E_{\mathrm{rest}}}{c^2}
\]

Using the conversion factor:

\[
\begin{aligned}
\Delta m
&=\frac{0.80000572}{931.5}\\
&=8.588\times10^{-4}\ \mathrm{u}
\end{aligned}
\]

The nucleus loses approximately:

\[
\Delta m=8.588\times10^{-4}\ \mathrm{u}
\]

Its actual rest mass has decreased, but its mass number remains 60.

Part (d): Determine the change in binding energy

The initial and final nuclei contain the same:

• 27 protons.
• 33 neutrons.

Therefore, the combined mass of their separate nucleons is unchanged.

The final nucleus has a smaller rest mass, meaning its mass defect is greater.

Since binding energy is related to mass defect, the final nucleus has a greater total binding energy.

The increase is:

\[
\Delta B=B_f-B_i
\]

\[
\Delta B=(m_i-m_f)c^2
\]

Therefore:

\[
\Delta B=0.80000572\ \mathrm{MeV}
\]

The binding energy increases by approximately \(0.800\ \mathrm{MeV}\).

Now evaluate the four claims.

Claim 1: The mass number must decrease because the nucleus loses mass.

Incorrect.

Mass number counts nucleons. It is not a measurement of the nucleus’s actual rest mass.

The nucleus still contains 27 protons and 33 neutrons, so:

\[
A=27+33=60
\]

Claim 2: The decrease in nuclear rest-mass energy must be exactly \(0.800\ \mathrm{MeV}\).

Not exactly.

The photon carries \(0.800\ \mathrm{MeV}\), but the recoiling nucleus also receives kinetic energy.

The actual decrease in nuclear rest-mass energy is approximately \(0.80000572\ \mathrm{MeV}\).

Ignoring recoil, the student’s answer would be an excellent HSC approximation.

Claim 3: Nuclear binding energy cannot change because the number of protons and neutrons stays constant.

Incorrect.

Binding energy depends on the energy of the nuclear state, not just the number of nucleons.

The same 60 nucleons can occupy different nuclear energy states.

The ground state is more tightly bound than the excited state, so gamma emission increases the total binding energy.

Claim 4: A massless photon cannot account for a decrease in nuclear rest mass.

Incorrect.

A photon has zero rest mass, but it carries energy and momentum.

The nucleus loses energy when the photon is emitted. Because energy contributes to rest mass, the nuclear rest mass decreases.

The photon does not need to possess rest mass to carry away energy.

This is the key distinction: zero photon rest mass does not mean zero photon energy.

07Where to take this next

When analysing any nuclear transition, keep two questions separate:

• Has the number of protons or neutrons changed? This determines whether the element or isotope has changed.
• Has the energy of the nucleus changed? This determines whether its rest mass and binding energy have changed.

Gamma emission answers the second question without changing the answer to the first.

Once you can separate nucleon number from actual nuclear mass, the next useful step is calculating binding energy per nucleon and comparing how tightly different nuclei are bound.