Atomic masses and nuclear composition

Weigh a single carbon atom on the most sensitive scale ever built and you would still be measuring almost nothing — a mass so small that kilograms are the wrong unit entirely. Physicists solved this by inventing a ruler sized for the job: the atomic mass unit, defined so that one atom of carbon-12 weighs exactly $12$ u. Every other atom in the periodic table is then measured against that one fixed post, and the numbers finally become manageable — a proton at $1.007276$ u, a neutron at $1.008665$ u, a uranium nucleus at a little over $238$ u. But mass is only half the story written on a nucleus. Buried inside every nuclide symbol $^A_Z X$ is a full census of what the nucleus contains: $Z$ protons and $N = A - Z$ neutrons, held together by the strongest force in nature over a distance smaller than an atom by a factor of ten thousand. Two nuclei can share the same $Z$ and differ only in $N$ — isotopes, chemically identical twins with different nuclear weight. They can share $A$ with different $Z$ , or share $N$ with different $Z$ and $A$ . The periodic table's neat integer-looking atomic masses are, on closer inspection, abundance-weighted averages over families of isotopes — which is exactly why chlorine's listed mass is $35.45$ , a number that belongs to no single atom of chlorine that has ever existed. This article builds that vocabulary and arithmetic from the ground up: the definition and SI value of the atomic mass unit, its energy equivalent through $E = mc^2$ , the nuclide notation $^A_Z X$ , and the three "iso-" families — isotopes, isobars, and isotones — that every later idea in this chapter (mass defect, binding energy, fission, fusion) is built out of.

Key takeaways

A nuclide is written $^A_Z X$ , with $Z$ = atomic number (protons = electrons), $N$ = neutron number, and $A = Z + N$ = mass number (total nucleons).
The atomic mass unit (u) is defined so $^{12}_6\text{C}$ has mass exactly $12$ u; in SI, $1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}$ .
Through $E = mc^2$ , $1\ \text{u}$ is equivalent to about $931.5\ \text{MeV}$ of energy — the conversion factor used throughout nuclear physics.
A proton has mass $m_p \approx 1.007276\ \text{u}$ and charge $+e$ ; a neutron has mass $m_n \approx 1.008665\ \text{u}$ and no charge — together, nucleons.
Isotopes share $Z$ (same element, different $A$ ); isobars share $A$ (different element, same total nucleon count); isotones share $N$ (different element, same neutron count).
The periodic-table atomic mass of an element is an abundance-weighted average over its naturally occurring isotopes, not the mass of any single atom.
Key idea: a nuclide's identity is fixed by the count Z protons and N = A minus Z neutrons written as A over Z X, its mass is measured in atomic mass units pinned to carbon-12 and convertible to energy through 1 u equals 931.5 MeV, and nuclides sharing Z, A, or N are called isotopes, isobars, or isotones respectively.

The core insight

A nuclide $^A_Z X$ packs $Z$ protons and $N = A-Z$ neutrons whose combined mass is measured in atomic mass units fixed by carbon-12 ( $1\ \text{u} = 1.660539\times10^{-27}\ \text{kg} \equiv 931.5\ \text{MeV}$ ), and nuclides that share $Z$ , $A$ , or $N$ with another are called isotopes, isobars, or isotones respectively.

Where it fits in the syllabus

This topic opens Unit IX's chapter "Nuclei" in CBSE Class 12 Physics, immediately after "Atoms," and is squarely within the 2025-26 prescribed scope: composition and size of the nucleus, along with the mass-energy relation used to express atomic and nuclear masses. It deliberately stays qualitative-to-moderate in depth — no derivation of nuclear force models, and (per the current CBSE rationalisation) no radioactive decay law is assumed or required here. Everything introduced — nuclide notation, the atomic mass unit, isotopes/isobars/isotones — is vocabulary and arithmetic that mass defect, binding energy, and fission/fusion (later topics in this chapter) will lean on directly, so getting it precise now pays off for the rest of the chapter.

The physical picture

Every atom's nucleus is a tightly packed cluster of only two particle types, protons and neutrons, collectively called nucleons. A proton carries charge $+e$ and mass $m_p \approx 1.007276\ \text{u}$ ; a neutron is electrically neutral with a very slightly larger mass, $m_n \approx 1.008665\ \text{u}$ . The standard shorthand for describing any nucleus is the nuclide symbol

^{A}_{Z}X

where: $X$ is the chemical symbol of the element, $Z$ is the atomic number (the number of protons, which also fixes the element's chemical identity and equals the number of electrons in the neutral atom), $N$ is the neutron number, and $A = Z + N$ is the mass number, the total count of nucleons. Reading the symbol is purely bookkeeping: the subscript tells you the proton count, the superscript tells you the total nucleon count, and subtracting the two gives the neutron count. For $^{197}_{79}\text{Au}$ , gold, $Z = 79$ protons, $A = 197$ nucleons in all, so $N = 197 - 79 = 118$ neutrons. Because nuclear masses are far too small to state usefully in kilograms, physicists use the atomic mass unit, defined by fixing one specific atom as the reference: the mass of a neutral $^{12}_6\text{C}$ atom is set to be exactly $12$ u, so $1$ u is exactly one-twelfth of that mass. In SI units,

1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}

where: this conversion lets any atomic or nuclear mass quoted in u be turned into kilograms for use in Newtonian formulas. Mass and energy are related by Einstein's relation

E = mc^2

where: $E$ is energy (joules), $m$ is mass (kg), and $c = 3\times10^8\ \text{m/s}$ is the speed of light in vacuum. Applying this to one atomic mass unit converts it into an energy equivalent of about $931.5$ MeV — a number used constantly to turn tiny mass differences into the large energies nuclear reactions release. Because it is so directly useful, physicists sometimes even quote nuclear masses in $\text{MeV}/c^2$ rather than in kilograms or u.

Figure

A labelled nucleus made of red proton circles and blue neutron circles clustered together beside the nuclide symbol A over Z X with arrows pointing to the mass number, atomic number, and element letter

Nuclides related to one another fall into three named families, and the naming is a mnemonic in disguise:

Isotopes share the same $Z$ (same element, same electron count, same chemistry) but different $A$ — for example hydrogen's three isotopes protium $^1_1\text{H}$ , deuterium $^2_1\text{H}$ , and tritium $^3_1\text{H}$ .
Isobars share the same $A$ but different $Z$ — for example $^3_1\text{H}$ and $^3_2\text{He}$ , both with $3$ total nucleons but a different split between protons and neutrons.
Isotones share the same $N$ but different $Z$ and $A$ — for example $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ , both with $N = 118$ neutrons.

The measured atomic mass listed in the periodic table for any element is not the mass of one atom of one isotope — it is the abundance-weighted average over every isotope that occurs naturally, which is precisely why so many listed atomic masses (chlorine at $35.45$ u, for instance) are not close to any whole number, even though every individual isotope's mass is close to a whole number of u.

Derivation ATurning one atomic mass unit into its energy equivalent

The goal is to derive the famous conversion factor $1\ \text{u} \approx 931.5\ \text{MeV}$ , starting only from the SI definition of the atomic mass unit and Einstein's mass-energy relation $E = mc^2$ , so that from this point on any mass quoted in u can be converted directly into an energy in MeV without repeating the unit-juggling each time. This single number is what makes it possible to read off, at a glance, how much energy a given mass difference represents in a nuclear process — the arithmetic backbone of mass defect and binding energy calculations later in the chapter.

Step 1. Start from the SI mass of one atomic mass unit, $m = 1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}$ , and apply $E = mc^2$ with $c = 2.998\times10^8\ \text{m/s}$ : $E = (1.660539\times10^{-27})\times(2.998\times10^8)^2$ .

Step 2. Square the speed of light first: $c^2 = (2.998\times10^8)^2 = 8.988\times10^{16}\ \text{m}^2/\text{s}^2$ .

Step 3. Multiply: $E = 1.660539\times10^{-27}\times8.988\times10^{16} = 1.4924\times10^{-10}\ \text{J}$ . This is the energy equivalent of $1$ u in joules — an extremely small number of joules, which is exactly why joules are the wrong unit for a single nuclear event.

Step 4. Convert joules to electron-volts using $1\ \text{eV} = 1.602\times10^{-19}\ \text{J}$ : $E = \dfrac{1.4924\times10^{-10}}{1.602\times10^{-19}} = 9.315\times10^{8}\ \text{eV} = 931.5\ \text{MeV}$ .

Step 5. State the reusable result: $1\ \text{u} = 931.5\ \text{MeV}/c^2$ as a mass, or equivalently $1\ \text{u} \to 931.5\ \text{MeV}$ of energy if fully converted. From here, any mass defect measured in u can be turned into a released energy in MeV simply by multiplying by $931.5$ — no need to repeat Steps 1 through 4 each time.

Worked examples

Board level - Average atomic mass of chlorine from its isotopes

Question: Naturally occurring chlorine consists of two isotopes: $^{35}\text{Cl}$ (mass $34.97$ u, abundance $75.77\%$ ) and $^{37}\text{Cl}$ (mass $36.97$ u, abundance $24.23\%$ ). Calculate the average atomic mass of chlorine, and state whether this number is the mass of any single chlorine atom.

Step 1. The average atomic mass is the abundance-weighted sum of the isotopic masses, each multiplied by its fractional abundance: $\bar{m} = (0.7577\times34.97) + (0.2423\times36.97)$ .

Step 2. Compute each term separately: $0.7577\times34.97 = 26.497$ u, and $0.2423\times36.97 = 8.960$ u.

Step 3. Add the two contributions: $\bar m = 26.497 + 8.960 = 35.457$ u.

Answer: $\bar m \approx \mathbf{35.46\ u}$ , matching the value printed for chlorine in the periodic table — and no, this is not the mass of any real chlorine atom, since every actual atom is either $^{35}\text{Cl}$ or $^{37}\text{Cl}$ ; the $35.46$ figure only describes the statistical mixture found in a natural sample.

Competitive level - Nuclide bookkeeping and the energy equivalent of a mass difference

Question: For the nuclide $^{197}_{79}\text{Au}$ (gold), find $Z$ , $N$ , and $A$ , give one isotope, one isobar, and one isotone example. Then, if a nuclear process involving gold released a mass difference of exactly $0.0025$ u, find the energy released in MeV.

Step 1. Read the nuclide symbol directly: $A = 197$ (mass number) and $Z = 79$ (atomic number, protons).

Step 2. Find the neutron number by subtraction: $N = A - Z = 197 - 79 = 118$ .

Step 3. Give the three related nuclides: an isotope of gold keeps $Z = 79$ with a different $A$ , e.g. $^{195}_{79}\text{Au}$ ; an isobar keeps $A = 197$ with a different $Z$ , e.g. $^{197}_{78}\text{Pt}$ (platinum); an isotone keeps $N = 118$ with different $Z$ and $A$ , e.g. $^{198}_{80}\text{Hg}$ , since $198 - 80 = 118$ matches gold's neutron count.

Step 4. Convert the mass difference to energy using the derived factor $1\ \text{u} = 931.5\ \text{MeV}$ : $E = 0.0025\times931.5 = 2.329\ \text{MeV}$ .

Answer: $Z = 79$ , $N = 118$ , $A = 197$ ; isotope $^{195}_{79}\text{Au}$ , isobar $^{197}_{78}\text{Pt}$ , isotone $^{198}_{80}\text{Hg}$ ; energy released $\approx \mathbf{2.33\ MeV}$ . Timing note: the nuclide bookkeeping (Steps 1-3) should take well under a minute once $A = Z + N$ is automatic; do not waste exam time hunting for isotope/isobar/isotone examples from memory — construct them on the spot from the definitions, exactly as done above, and keep the $1\ \text{u} = 931.5\ \text{MeV}$ conversion memorised cold since it recurs in nearly every later numerical in this chapter.

Common traps

Confusing mass number $A$ (total nucleons) with atomic number $Z$ (protons only) — the subscript is $Z$ , the superscript is $A$ , and $N = A - Z$ always gives the neutron count; never read $A$ as if it were the number of protons.
Mixing up isotopes, isobars, and isotones under exam pressure — use the mnemonic that isoTOpes match $Z$ (same element, so same position on the periodic Table), isoBArs match $A$ (same total nucleon count, Add up the same), and isoTONes match $N$ (same Neutron count).
Believing isotopes have different chemical properties — they do not; identical $Z$ means identical electron count and hence identical chemistry, even though nuclear mass and nuclear stability differ between isotopes.
Treating the periodic-table atomic mass as the mass of a single atom, then being confused when it is not a whole number — it is an abundance-weighted average over all naturally occurring isotopes of that element.
Forgetting to convert $c$ or the mass to consistent SI units before applying $E = mc^2$ , which silently produces an answer that is many orders of magnitude wrong; always quote $c$ in m/s and mass in kg before multiplying, or use the ready-made $1\ \text{u} = 931.5\ \text{MeV}$ shortcut instead of re-deriving it.

You should be able to…

Write the standard nuclide notation $^A_Z X$ and extract $Z$ , $N$ , and $A$ from it, or reconstruct the notation given the three numbers.
State the SI definition of the atomic mass unit and its numerical value, $1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}$ .
Derive or recall the energy equivalent $1\ \text{u} = 931.5\ \text{MeV}$ using $E = mc^2$ , and apply it to convert a mass difference into an energy in MeV.
Distinguish isotopes, isobars, and isotones with a correctly constructed example of each, on demand.
Compute a weighted-average atomic mass from a set of isotopic masses and their fractional abundances.
Explain why isotopes of the same element are chemically identical despite having different masses.

Ask the tutor: if you have a nuclide's $Z$ , $A$ , or a mass difference in u and want a check on the isotope/isobar/isotone classification or the MeV conversion, ask the AI tutor to work through your specific numbers step by step.

Why not just use kilograms for atomic masses like we do for everything else?

You could, but the numbers become unwieldy and error-prone: a carbon atom's mass in kilograms is about $1.99\times10^{-26}\ \text{kg}$ , and comparing two nearby isotopes means comparing tiny differences between two already-tiny numbers with many decimal places. Defining $1$ u as one-twelfth of a carbon-12 atom's mass turns every atomic mass into a number close to a small integer (hydrogen near $1$ , oxygen near $16$ , uranium near $238$ ), which is far easier to read, compare, and remember, while still converting cleanly to kilograms or MeV whenever the SI unit is actually needed.

Why is carbon-12 specifically the reference, and not hydrogen or oxygen?

Historically both hydrogen and oxygen were tried as the reference at different times, but carbon-12 was adopted internationally because it is a stable, abundant, easy-to-purify isotope, and choosing it as exactly $12$ u keeps most other atomic masses close to whole numbers without needing an awkward fractional reference point. It is a convention, not a law of physics — any single fixed reference mass would work in principle, but carbon-12 is the one the world agreed on, so every u value you look up anywhere is measured against that same post.

If neutrons have no charge, how do we know they are even inside the nucleus?

Neutrons were inferred and then directly detected because a nucleus's mass is far larger than what its charge and known proton count alone would predict, and because certain nuclear reactions clearly involve an electrically neutral particle carrying away mass and energy. James Chadwick confirmed the neutron experimentally in 1932 by observing how an unknown neutral radiation, when it struck light nuclei like those in paraffin wax, knocked protons forward with an amount of energy that only made sense if the incoming particle had almost the same mass as a proton but zero charge.

Is radioactivity part of this topic, since old textbooks always mention it here?

No — under the current CBSE 2025-26 rationalisation, the law of radioactive decay, half-life, mean life, and activity are not listed as theory content for this chapter, even though many older resources and previous editions of the textbook include them right after nuclear composition. This article deliberately stays with composition, atomic mass units, and the isotope/isobar/isotone families, which are the parts that remain in the prescribed 2025-26 syllabus; do not prepare radioactive decay formulas expecting them from this topic.