How to Calculate Binding Energy per Nucleon from Nuclear Mass

A data sheet gives the nuclear mass of oxygen-16 as \(15.990526\ \mathrm{u}\). Can you calculate its binding energy per nucleon by simply dividing this mass by 16?

No. That is just the mass of the bound nucleus per nucleon. To calculate binding energy per nucleon, you first need to calculate the combined mass of eight separate protons and eight separate neutrons. You then compare this with the measured nuclear mass, convert the difference into energy, and divide by 16.

The method is straightforward once you know which masses to use. The main difficulty in HSC questions is often not the arithmetic. It is recognising whether the question gives you a nuclear mass or an atomic mass, and making sure you count the protons and neutrons correctly.

01The formula: calculating binding energy per nucleon

Start with the isotope notation:

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

The two numbers tell you exactly how many particles make up the nucleus:

• \(A\) is the mass number, or total number of nucleons.
• \(Z\) is the atomic number, or number of protons.
• \(N=A-Z\) is the number of neutrons.

For a nucleus with \(Z\) protons and \(N\) neutrons, the combined mass of its separate nucleons is:

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

Here:

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

The actual nuclear mass is slightly smaller than this combined mass.

The difference is called the mass defect:

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

Once you have calculated the mass defect, convert it into binding energy using mass-energy equivalence.

If your masses are in atomic mass units (u), use:

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

Therefore:

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

Finally, divide the total binding energy by the mass number:

\[ \boxed{ \frac{E_b}{A} = \frac{\Delta m\times931.5}{A} } \]

The result is measured in MeV per nucleon.

This is the calculation you need when a question supplies the nuclear mass directly.

The conversion factor \(931.5\ \mathrm{MeV/u}\) is the rounded energy equivalent of one atomic mass unit. The more precise recommended value is available from NIST’s CODATA constants.

What does the calculation actually compare?

Imagine taking a nucleus apart until you have a collection of completely separate protons and neutrons.

The separate particles have a greater combined rest mass than the original bound nucleus. The difference corresponds to the energy released when those particles formed the nucleus.

The mass defect is converted into the total binding energy. Dividing that energy by the number of nucleons gives the average binding energy per nucleon.

A larger binding energy per nucleon generally indicates that a nucleus is more tightly bound on average.

For a fuller explanation of why the mass difference exists, see Mass defect: how missing mass becomes nuclear binding energy.

02Step 1: Count the protons and neutrons correctly

Before using your calculator, check the isotope notation.

Consider oxygen-16:

\[ {}^{16}_{8}\mathrm{O} \]

The bottom number tells you there are eight protons.

The top number tells you there are 16 nucleons altogether.

Therefore:

\[ N=A-Z=16-8=8 \]

The nucleus contains:

• 8 protons
• 8 neutrons
• 16 nucleons in total

This is important because using the wrong number of protons or neutrons changes the calculated mass defect.

Charge and nucleon number conservation

You also need to recognise how proton and nucleon numbers behave in nuclear reactions.

Two quantities are conserved:

1. Total nucleon number (mass number, \(A\)).
2. Total nuclear charge (proton number, \(Z\)).

For example:

\[ {}^{19}_{9}\mathrm{F} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{23}_{11}\mathrm{Na} + \gamma \]

Check the nucleon numbers:

\[ 19+4=23 \]

Check the proton numbers:

\[ 9+2=11 \]

Both balance. The gamma photon has zero nucleon number and zero electric charge.

This is useful when nuclear masses are supplied as part of a reaction. You must know which particles are present before calculating the mass difference.

03Where do measured nuclear masses come from?

A nucleus is far too small to place on an ordinary balance. Instead, scientists determine atomic and nuclear masses using measurements of particle motion.

One common experimental technique is magnetic-sector mass spectrometry.

The apparatus works in a sequence:

1. A sample is converted into a gas and its atoms are ionised.
2. An electric field accelerates the ions through a known potential difference.
3. The ions enter a region containing a magnetic field, perpendicular to their motion.
4. The magnetic force bends their paths.
5. A detector records where the ions arrive.

What should the detector show?

For ions with the same charge, accelerated through the same potential difference and entering the same magnetic field, heavier ions follow paths with larger radii.

The relationship is:

\[ r=\frac{1}{B}\sqrt{\frac{2mV}{q}} \]

where:

• \(r\) is the radius of the curved path, in metres.
• \(B\) is the magnetic flux density, in teslas.
• \(m\) is the ion mass, in kilograms.
• \(V\) is the accelerating potential difference, in volts.
• \(q\) is the magnitude of the ion’s charge, in coulombs.

For a quick scale estimate, compare singly charged ions with masses of 16 u and 18 u.

At the same accelerating voltage and magnetic field:

\[ \frac{r_{18}}{r_{16}} = \sqrt{\frac{18}{16}} \]

\[ \frac{r_{18}}{r_{16}}\approx1.061 \]

The heavier ion follows a path with a radius approximately 6.1% larger.

The detector therefore distinguishes isotopes through their different mass-to-charge ratios.

There is one important limitation: a mass spectrometer measures the mass-to-charge ratio of an ion, not the mass of a bare nucleus directly. The ion may still contain electrons. Its charge state must be known, and the appropriate electron correction must be made when determining the atomic or nuclear mass.

This is why the distinction between nuclear mass and atomic mass matters in your calculation.

04Step 2: Identify whether you have a nuclear mass or an atomic mass

This is one of the most important decisions in a binding energy calculation.

A nuclear mass and an atomic mass are not the same thing.

A neutral atom containing \(Z\) protons also contains \(Z\) electrons.

The electron masses are small, but they are not zero. Using the wrong mass combination can noticeably change your answer.

Case 1: The question gives you nuclear mass

If the question explicitly gives the nuclear mass, use the proton and neutron masses directly.

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

There are no electrons to account for.

Case 2: The question gives you atomic mass

If the question gives the mass of a neutral atom, you need to account for its electrons.

A convenient method is to use the mass of a neutral hydrogen atom instead of the mass of a proton.

The reason is that a neutral hydrogen atom contains one proton and one electron.

Therefore, the mass of \(Z\) hydrogen atoms includes the same number of electron rest masses as a neutral atom containing \(Z\) protons.

The appropriate expression becomes:

\[ \Delta m=Zm_H+Nm_n-m_{\text{atom}} \]

where \(m_H\) is the mass of one neutral hydrogen atom.

The electron rest-mass contributions cancel. At HSC calculation precision, the small differences in electronic binding energies are neglected.

A useful decision table

Never combine proton masses with neutral atomic masses without correcting for the electrons.

Also, be careful with periodic tables. The average atomic mass of an element is generally a weighted average of its naturally occurring isotopes. It is not necessarily the mass of the particular isotope in your question.

07Binding energy versus energy released in a nuclear reaction

There is another distinction worth getting right.

Binding energy and energy released in a nuclear reaction are related, but they are not interchangeable.

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

Equivalently, it is the energy released when those free nucleons form the bound nucleus.

Reaction energy, usually represented by \(Q\), is the energy released or absorbed when a particular nuclear reaction takes place.

For a reaction involving initial and final masses:

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

When the mass difference is in u:

\[ Q=(m_{\text{initial}}-m_{\text{final}}) \times931.5 \]

The sign tells you what happens:

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

The mass comparison must include all relevant initial and final particles, with consistent mass definitions.

A reaction’s released energy is generally not equal to the binding energy of its product nucleus. It depends on the difference between the total binding energies of all the reactants and products.

For a more detailed explanation of the mass-to-energy conversion, see Nuclear Binding Energy: Convert Mass Defect with E = mc².

08Questions and solutions

The following questions build from a straightforward nuclear mass calculation to more demanding applications involving atomic masses, conservation laws, and reaction energy.

Question 1

An alpha particle is a helium-4 nucleus with two protons and two neutrons.

Its measured nuclear mass is \(4.001506\ \mathrm{u}\).

Use:

• Proton mass: \(1.007276\ \mathrm{u}\)
• Neutron mass: \(1.008665\ \mathrm{u}\)

Calculate:

(a) The mass defect of the alpha particle.

(b) Its total binding energy in MeV.

(c) Its binding energy per nucleon.

Solution 1

The total binding energy is approximately \(28.30\ \mathrm{MeV}\), giving a binding energy per nucleon of \(7.07\ \mathrm{MeV/nucleon}\).

Step 1: Identify the number of protons and neutrons.

For helium-4:

\[ A=4,\qquad Z=2 \]

Therefore:

\[ N=4-2=2 \]

Step 2: Calculate the combined mass of the separate nucleons.

\[ \begin{aligned} m_{\text{separate}} &=2m_p+2m_n\\ &=2(1.007276)+2(1.008665)\\ &=2.014552+2.017330\\ &=4.031882\ \mathrm{u} \end{aligned} \]

Step 3: Calculate the mass defect.

\[ \begin{aligned} \Delta m &=4.031882-4.001506\\ &=0.030376\ \mathrm{u} \end{aligned} \]

Step 4: Calculate total binding energy.

\[ \begin{aligned} E_b &=0.030376\times931.5\\ &=28.295\ \mathrm{MeV} \end{aligned} \]

Step 5: Divide by the mass number.

\[ \frac{E_b}{A} = \frac{28.295}{4} \]

\[ \boxed{ E_b/A=7.07\ \mathrm{MeV/nucleon} } \]

The alpha particle has a total binding energy of approximately \(28.30\ \mathrm{MeV}\). Its average binding energy per nucleon is approximately \(7.07\ \mathrm{MeV}\).

Question 2

A neutral magnesium-24 atom has an atomic mass of \(23.985042\ \mathrm{u}\).

Use the following illustrative rounded values:

(a) Calculate the binding energy per nucleon of magnesium-24.

(b) A student calculates the mass defect using:

\[ \Delta m=12m_p+12m_n-m_{\text{atom}} \]

Explain why this method is incorrect and calculate how much it changes the resulting binding energy per nucleon.

Solution 2

The correct binding energy per nucleon is approximately \(8.26\ \mathrm{MeV/nucleon}\). The student’s method underestimates it because it fails to account for the 12 electrons included in the atomic mass.

Step 1: Determine the number of protons and neutrons.

Magnesium-24 is:

\[ {}^{24}_{12}\mathrm{Mg} \]

Therefore:

\[ Z=12,\qquad N=24-12=12 \]

Step 2: Select the correct mass expression.

The supplied mass is an atomic mass.

Therefore:

\[ \Delta m=12m_H+12m_n-m_{\text{atom}} \]

Step 3: Calculate the mass defect.

\[ \begin{aligned} m_{\text{reference}} &=12(1.007825)+12(1.008665)\\ &=12.093900+12.103980\\ &=24.197880\ \mathrm{u} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta m &=24.197880-23.985042\\ &=0.212838\ \mathrm{u} \end{aligned} \]

Step 4: Calculate the total binding energy.

\[ \begin{aligned} E_b &=0.212838\times931.5\\ &=198.259\ \mathrm{MeV} \end{aligned} \]

Step 5: Calculate binding energy per nucleon.

\[ \frac{E_b}{A} = \frac{198.259}{24} \]

\[ \boxed{ E_b/A=8.26\ \mathrm{MeV/nucleon} } \]

Step 6: Identify the student’s error.

The student has used proton masses but subtracted a neutral atomic mass.

The atomic mass includes 12 electrons, but the 12 proton masses do not.

The mass defect is therefore underestimated by:

\[ \begin{aligned} \Delta m_{\text{error}} &=12m_e\\ &=12(0.000549)\\ &=0.006588\ \mathrm{u} \end{aligned} \]

The resulting error in total binding energy is:

\[ \Delta E=0.006588\times931.5 \]

\[ \Delta E=6.137\ \mathrm{MeV} \]

The error per nucleon is:

\[ \frac{6.137}{24} = 0.256\ \mathrm{MeV/nucleon} \]

The incorrect method would give approximately \(8.01\ \mathrm{MeV/nucleon}\), rather than \(8.26\ \mathrm{MeV/nucleon}\).

The important lesson is that the masses on both sides of your subtraction must account for the same particles.

Question 3

Consider the nuclear reaction:

\[ {}^{19}_{9}\mathrm{F} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{23}_{11}\mathrm{Na} + \gamma \]

The following neutral atomic masses are provided:

Also use:

• Hydrogen atom mass: \(1.007825\ \mathrm{u}\)
• Neutron mass: \(1.008665\ \mathrm{u}\)

(a) Verify that the nuclear reaction conserves charge and nucleon number.

(b) Calculate the energy released by the reaction.

(c) Calculate the binding energy per nucleon of sodium-23.

(d) Explain why the energy released in part (b) is not the binding energy per nucleon of sodium-23.

Solution 3

The reaction releases approximately \(10.47\ \mathrm{MeV}\), while sodium-23 has a binding energy per nucleon of approximately \(8.11\ \mathrm{MeV/nucleon}\). These are different quantities.

Step 1: Check nucleon number conservation.

On the reactant side:

\[ A=19+4=23 \]

On the product side:

\[ A=23 \]

Nucleon number is conserved.

Step 2: Check charge conservation.

On the reactant side:

\[ Z=9+2=11 \]

On the product side:

\[ Z=11 \]

Charge is also conserved.

Step 3: Calculate the energy released.

Using the neutral atomic masses is valid as an electron-mass bookkeeping method because the total proton number is the same on both sides.

The initial mass is:

\[ \begin{aligned} m_i &=18.998403+4.002603\\ &=23.001006\ \mathrm{u} \end{aligned} \]

The final mass is:

\[ m_f=22.989770\ \mathrm{u} \]

Therefore:

\[ \begin{aligned} \Delta m &=m_i-m_f\\ &=23.001006-22.989770\\ &=0.011236\ \mathrm{u} \end{aligned} \]

Calculate the released energy:

\[ \begin{aligned} Q &=0.011236\times931.5\\ &=10.466\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \boxed{Q=10.47\ \mathrm{MeV}} \]

The positive value means the reaction releases energy.

Step 4: Calculate the binding energy of sodium-23.

First, identify its composition:

\[ A=23,\qquad Z=11 \]

\[ N=23-11=12 \]

Because the supplied sodium mass is an atomic mass, use the hydrogen atom mass.

\[ \begin{aligned} m_{\text{reference}} &=11m_H+12m_n\\ &=11(1.007825)+12(1.008665)\\ &=11.086075+12.103980\\ &=23.190055\ \mathrm{u} \end{aligned} \]

Calculate the mass defect:

\[ \begin{aligned} \Delta m &=23.190055-22.989770\\ &=0.200285\ \mathrm{u} \end{aligned} \]

Convert this into total binding energy:

\[ \begin{aligned} E_b &=0.200285\times931.5\\ &=186.565\ \mathrm{MeV} \end{aligned} \]

Divide by 23 nucleons:

\[ \frac{E_b}{A} = \frac{186.565}{23} \]

\[ \boxed{ E_b/A=8.11\ \mathrm{MeV/nucleon} } \]

Step 5: Explain the difference.

The \(10.47\ \mathrm{MeV}\) is the energy released by the entire reaction.

The \(8.11\ \mathrm{MeV/nucleon}\) is the average binding energy of the sodium-23 nucleus.

The reaction energy depends on the difference between the combined binding energies of the reactants and the product. It is not simply the binding energy of the product.

In this reaction, the product has a total binding energy greater than the combined initial binding energies by approximately \(10.47\ \mathrm{MeV}\).

Question 4

Two neutral atoms have the following masses:

Use:

• Hydrogen atom mass: \(1.007825\ \mathrm{u}\)
• Neutron mass: \(1.008665\ \mathrm{u}\)

(a) Calculate the total binding energy of each nucleus.

(b) Calculate the binding energy per nucleon of each nucleus.

(c) A student claims that uranium-238 must be more tightly bound because its total binding energy is much greater than that of iron-56. Assess this claim.

Solution 4

Iron-56 has the greater binding energy per nucleon, approximately \(8.79\ \mathrm{MeV/nucleon}\), compared with \(7.57\ \mathrm{MeV/nucleon}\) for uranium-238. The student’s conclusion confuses total binding energy with average binding energy per nucleon.

Step 1: Identify the composition of each nucleus.

For iron-56:

\[ Z=26,\qquad N=56-26=30 \]

For uranium-238:

\[ Z=92,\qquad N=238-92=146 \]

Step 2: Calculate the mass defect of iron-56.

Using atomic mass:

\[ \Delta m_{\mathrm{Fe}} = 26m_H+30m_n-m_{\mathrm{atom}} \]

Substitute:

\[ \begin{aligned} m_{\text{reference}} &=26(1.007825)+30(1.008665)\\ &=26.203450+30.259950\\ &=56.463400\ \mathrm{u} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta m_{\mathrm{Fe}} &=56.463400-55.934936\\ &=0.528464\ \mathrm{u} \end{aligned} \]

Step 3: Calculate the total binding energy of iron-56.

\[ \begin{aligned} E_{b,\mathrm{Fe}} &=0.528464\times931.5\\ &=492.264\ \mathrm{MeV} \end{aligned} \]

Step 4: Calculate the binding energy per nucleon of iron-56.

\[ \frac{E_{b,\mathrm{Fe}}}{A} = \frac{492.264}{56} \]

\[ \boxed{ E_{b,\mathrm{Fe}}/A =8.79\ \mathrm{MeV/nucleon} } \]

Step 5: Calculate the mass defect of uranium-238.

\[ \Delta m_{\mathrm{U}} = 92m_H+146m_n-m_{\mathrm{atom}} \]

Substitute:

\[ \begin{aligned} m_{\text{reference}} &=92(1.007825)+146(1.008665)\\ &=92.719900+147.265090\\ &=239.984990\ \mathrm{u} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta m_{\mathrm{U}} &=239.984990-238.050788\\ &=1.934202\ \mathrm{u} \end{aligned} \]

Step 6: Calculate the total binding energy of uranium-238.

\[ \begin{aligned} E_{b,\mathrm{U}} &=1.934202\times931.5\\ &=1801.709\ \mathrm{MeV} \end{aligned} \]

Step 7: Calculate the binding energy per nucleon.

\[ \frac{E_{b,\mathrm{U}}}{A} = \frac{1801.709}{238} \]

\[ \boxed{ E_{b,\mathrm{U}}/A =7.57\ \mathrm{MeV/nucleon} } \]

Step 8: Assess the student’s claim.

The student is correct that uranium-238 has a much larger total binding energy:

• Iron-56: approximately \(492.3\ \mathrm{MeV}\).
• Uranium-238: approximately \(1801.7\ \mathrm{MeV}\).

However, uranium-238 also contains considerably more nucleons.

When we compare the average binding energy per nucleon, iron-56 has the greater value.

Therefore, iron-56 is more tightly bound on average in this comparison.

This illustrates why total binding energy alone is not a useful way to compare nuclei of substantially different sizes. Also, binding energy per nucleon is an average measure, not a direct prediction of an isotope’s half-life.

Question 5

A neutral neon-20 atom captures a free neutron and forms neon-21:

\[ {}^{20}_{10}\mathrm{Ne} + {}^{1}_{0}n \rightarrow {}^{21}_{10}\mathrm{Ne} + \gamma \]

The following illustrative rounded data are supplied:

Assume electronic binding-energy differences are negligible. The neutron is initially at rest, and treat the gamma photon as carrying approximately all the released energy, neglecting the small nuclear recoil.

(a) Verify that the reaction conserves nucleon number and charge.

(b) Calculate the binding energy per nucleon of neon-20.

(c) Calculate the binding energy per nucleon of neon-21.

(d) Calculate the energy released by neutron capture.

(e) A student argues:

“The reaction releases energy, so neon-21 must have a greater binding energy per nucleon than neon-20.”

Determine whether the student is correct. Explain your reasoning using your calculated values.

Solution 5

The reaction releases approximately \(6.76\ \mathrm{MeV}\), but the binding energy per nucleon decreases from approximately \(8.03\ \mathrm{MeV/nucleon}\) to \(7.97\ \mathrm{MeV/nucleon}\). The student is incorrect.

This is possible because total binding energy and average binding energy per nucleon measure different things.

Step 1: Check conservation laws.

For nucleon number:

\[ 20+1=21 \]

For charge:

\[ 10+0=10 \]

Both quantities are conserved.

Step 2: Calculate the binding energy of neon-20.

Neon-20 contains:

\[ Z=10,\qquad N=20-10=10 \]

Because the supplied mass is an atomic mass, use the hydrogen atom mass.

\[ \begin{aligned} m_{\text{reference}} &=10m_H+10m_n\\ &=10(1.007825)+10(1.008665)\\ &=10.078250+10.086650\\ &=20.164900\ \mathrm{u} \end{aligned} \]

Calculate the mass defect:

\[ \begin{aligned} \Delta m_{20} &=20.164900-19.992440\\ &=0.172460\ \mathrm{u} \end{aligned} \]

Convert to total binding energy:

\[ \begin{aligned} E_{b,20} &=0.172460\times931.5\\ &=160.646\ \mathrm{MeV} \end{aligned} \]

Divide by 20 nucleons:

\[ \frac{E_{b,20}}{A} = \frac{160.646}{20} \]

\[ \boxed{ E_{b,20}/A =8.03\ \mathrm{MeV/nucleon} } \]

Step 3: Calculate the binding energy of neon-21.

Neon-21 contains:

\[ Z=10,\qquad N=21-10=11 \]

The combined reference mass is:

\[ \begin{aligned} m_{\text{reference}} &=10m_H+11m_n\\ &=10(1.007825)+11(1.008665)\\ &=10.078250+11.095315\\ &=21.173565\ \mathrm{u} \end{aligned} \]

Calculate the mass defect:

\[ \begin{aligned} \Delta m_{21} &=21.173565-20.993847\\ &=0.179718\ \mathrm{u} \end{aligned} \]

Convert to total binding energy:

\[ \begin{aligned} E_{b,21} &=0.179718\times931.5\\ &=167.407\ \mathrm{MeV} \end{aligned} \]

Divide by 21 nucleons:

\[ \frac{E_{b,21}}{A} = \frac{167.407}{21} \]

\[ \boxed{ E_{b,21}/A =7.97\ \mathrm{MeV/nucleon} } \]

Step 4: Calculate the energy released by neutron capture.

The initial mass is:

\[ \begin{aligned} m_i &=19.992440+1.008665\\ &=21.001105\ \mathrm{u} \end{aligned} \]

The final mass is:

\[ m_f=20.993847\ \mathrm{u} \]

Therefore:

\[ \begin{aligned} \Delta m &=21.001105-20.993847\\ &=0.007258\ \mathrm{u} \end{aligned} \]

Convert this mass difference into energy:

\[ \begin{aligned} Q &=0.007258\times931.5\\ &=6.761\ \mathrm{MeV} \end{aligned} \]

Thus:

\[ \boxed{Q=6.76\ \mathrm{MeV}} \]

The reaction is exothermic because the initial total mass is greater than the final total mass.

Step 5: Explain why the student’s conclusion is incorrect.

The tempting argument is that an exothermic reaction must produce a nucleus with a greater binding energy per nucleon.

However, the reaction only tells us that the total binding energy increases.

Compare the two quantities:

The total binding energy increases by:

\[ 167.407-160.646 = 6.761\ \mathrm{MeV} \]

This is exactly the energy released in the neutron-capture reaction.

However, the original average was approximately \(8.03\ \mathrm{MeV}\) per nucleon.

The extra neutron increases the total binding energy by only \(6.76\ \mathrm{MeV}\), which is less than the original average of \(8.03\ \mathrm{MeV}\).

Adding this neutron therefore increases the total binding energy but decreases the average binding energy per nucleon.

The student has confused an increase in the total with an increase in the average.

This is an important distinction when interpreting nuclear binding energy calculations.

09What to remember in an HSC calculation

When a question asks you to calculate binding energy per nucleon from nuclear mass, follow this sequence:

1. Identify the isotope and read off \(A\) and \(Z\).
2. Calculate the number of neutrons using \(N=A-Z\).
3. Identify whether the supplied mass is a nuclear mass or a neutral atomic mass.
4. Calculate the combined reference mass using the correct proton, neutron, or hydrogen atom masses.
5. Subtract the measured mass to obtain the mass defect.
6. Multiply the mass defect in u by \(931.5\) to obtain total binding energy in MeV.
7. Divide the total binding energy by \(A\).

Keep the following distinctions clear:

Once you can calculate binding energy per nucleon reliably, the next useful step is interpreting the nuclear binding energy curve. It explains why fusion of suitable light nuclei and fission of suitable heavy nuclei can release energy, and why the total mass difference of a particular reaction must still be calculated separately.