Mass Defect Calculator
Calculate nuclear mass defect and binding energy from atomic composition. Find binding energy per nucleon with step-by-step solutions.
Mass Defect Calculator
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Formula: Mass Defect = (Z x mp + N x mn + Z x me) - M_atom
Worked example โ Mass defect = 0.0304 amu, BE = 28.3 MeV, BE/A = 7.07 MeV
Formula
Mass Defect = (Z x mp + N x mn + Z x me) - M_atom
The mass defect is the difference between the sum of individual particle masses (protons, neutrons, electrons) and the measured atomic mass. Binding energy equals the mass defect times 931.494 MeV/amu.
Worked Examples
Example 1: Helium-4 Mass Defect
Problem:Calculate the mass defect and binding energy of He-4 (2 protons, 2 neutrons, atomic mass 4.002602 amu).
Solution:Protons: 2 x 1.007276 = 2.014552 amu Neutrons: 2 x 1.008665 = 2.017330 amu Electrons: 2 x 0.000549 = 0.001098 amu Total = 4.032980 amu Mass defect = 4.032980 - 4.002602 = 0.030378 amu BE = 0.030378 x 931.494 = 28.296 MeV
Result:Mass defect = 0.0304 amu, BE = 28.3 MeV, BE/A = 7.07 MeV
Example 2: Iron-56 Mass Defect
Problem:Calculate for Fe-56 (26 protons, 30 neutrons, atomic mass 55.934936 amu).
Solution:Protons: 26 x 1.007276 = 26.189176 amu Neutrons: 30 x 1.008665 = 30.259950 amu Electrons: 26 x 0.000549 = 0.014274 amu Total = 56.463400 amu Mass defect = 56.463400 - 55.934936 = 0.528464 amu BE = 492.26 MeV, BE/A = 8.79 MeV
Result:Mass defect = 0.5285 amu, BE = 492.3 MeV, BE/A = 8.79 MeV
Frequently Asked Questions
What is mass defect?
Mass defect is the difference between the total mass of individual protons, neutrons, and electrons that make up an atom and the actual measured mass of the atom. The actual atomic mass is always less than the sum of its parts because some mass is converted to energy that holds the nucleus together (binding energy). This relationship is described by Einstein equation E = mc^2. A larger mass defect means stronger nuclear binding and a more stable nucleus.
What is nuclear binding energy?
Nuclear binding energy is the energy required to completely disassemble a nucleus into its individual protons and neutrons. It equals the mass defect multiplied by c^2, using the conversion factor 931.494 MeV per amu. Binding energy per nucleon peaks at iron-56 (about 8.8 MeV per nucleon), which is why iron is the most stable element. Elements lighter than iron can release energy through fusion, while heavier elements release energy through fission.
Why is binding energy per nucleon important?
Binding energy per nucleon is the key indicator of nuclear stability. The curve of binding energy per nucleon versus mass number rises steeply for light elements, peaks near iron-56, and gradually decreases for heavier elements. This shape explains why nuclear fusion of light elements (like hydrogen in stars) and fission of heavy elements (like uranium in reactors) both release energy. The difference in binding energy per nucleon between reactants and products determines the energy released.
How does mass defect relate to E = mc^2?
Einstein mass-energy equivalence E = mc^2 directly explains mass defect. When nucleons combine to form a nucleus, the binding energy released causes a measurable decrease in mass. For example, a helium-4 nucleus has a mass defect of 0.030 amu, corresponding to 28.3 MeV of binding energy. This energy was released when the nucleus formed and would need to be supplied to break it apart. The conversion factor 1 amu = 931.494 MeV/c^2 is derived from E = mc^2.
Why are electron masses included in the constituent total?
Tabulated atomic masses are for the neutral atom, so they already contain Z electrons. To subtract like from like, the constituent side must include those electrons too, at 0.000549 u each, alongside Z protons and N neutrons. Omitting them inflates the mass defect by about 0.0005 u per proton, which for uranium-238 is roughly 0.05 u or about 47 MeV of spurious binding energy. The alternative is to use bare nuclear masses on both sides, but evaluated tables almost always publish atomic masses, so counting the electrons is the safer convention.
What is the difference between mass number and atomic mass?
Mass number A is simply the count of nucleons, Z + N, and is always a whole number with no units. Atomic mass is the measured mass of the atom in unified atomic mass units, where carbon-12 is defined as exactly 12 u. The two are close but never equal except for carbon-12, and the gap is precisely what this calculation exploits. Helium-4 has A = 4 but an atomic mass of 4.002602 u; iron-56 has A = 56 but a mass of 55.934936 u, lighter than its mass number because of strong binding.
How do nuclear magic numbers show up in binding energy?
Protons and neutrons occupy quantised shells much like atomic electrons, and filling a shell completely produces extra stability. The closed-shell counts are 2, 8, 20, 28, 50, 82 and 126, and nuclei at those values sit measurably above the smooth binding energy trend. Helium-4 is doubly magic with two protons and two neutrons, which is why it already reaches about 7.07 MeV per nucleon despite being so light, and lead-208 with 82 protons and 126 neutrons is the heaviest stable doubly magic nucleus. These bumps explain anomalies a smooth curve cannot.
If iron-56 is called the most stable nucleus, why is nickel-62 sometimes cited instead?
Nickel-62 has the highest binding energy per nucleon of any nuclide, around 8.79 MeV, marginally ahead of iron-58 and iron-56. The three are so close that rounding decides which appears at the top of a table. Iron-56 nevertheless dominates in nature because stellar nucleosynthesis and the photodisintegration equilibrium in massive stars favour nickel-56, which then decays through cobalt-56 to iron-56. So nickel-62 wins on binding energy per nucleon while iron-56 wins on cosmic abundance, and both statements are correct in their own terms.
How is mass defect used to find the Q value of a nuclear reaction?
The Q value is the energy released, equal to the total mass of the reactants minus the total mass of the products, converted with 931.494 MeV per u. A positive Q means the reaction is exoergic and the missing mass appears as kinetic energy of the products. For the Cockcroft-Walton reaction, lithium-7 at 7.016004 u plus hydrogen-1 at 1.007825 u totals 8.023829 u, while two helium-4 nuclei total 8.005204 u; the difference of 0.018625 u corresponds to about 17.3 MeV. Using atomic masses throughout keeps the electron count balanced automatically.
How does the semi-empirical mass formula predict binding energy without measured masses?
The liquid-drop model of von Weizsaecker builds binding energy from competing terms. A volume term proportional to A rewards every nucleon for its nearest neighbours; a surface term subtracts a contribution proportional to A^(2/3) because surface nucleons have fewer neighbours; a Coulomb term subtracts proton-proton repulsion growing as Z^2/A^(1/3); an asymmetry term penalises any imbalance between N and Z; and a pairing term adds stability when both counts are even. It reproduces the overall curve well but misses shell closures, which is why magic-number nuclei still stand out.
References
Background & Theory
History
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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