Atomic number Z, mass number A, isotopes, isobars, isotones: Visual walkthrough

This page is the visual companion to the parent topic.

Step 1 — Draw the two things a nucleus is made of

WHAT: Before any letter or formula, we picture the nucleus as a small bag holding two kinds of ball: protons (we colour them coral) and neutrons (we colour them mint). Nothing else lives in the bag.

WHY: Every number we are about to invent is just counting balls in this bag. If you can see the bag, every formula becomes "count this, count that." A symbol you cannot picture is a symbol you cannot trust.

PICTURE: Look at the bag below. The coral balls carry a little "+"; the mint balls are blank (no charge). Around the bag, tiny slate dots orbit — those are electrons, far away and almost weightless. We are deliberately putting them outside to remind ourselves they don't add to the mass count.

Figure — Atomic number Z, mass number A, isotopes, isobars, isotones

Step 2 — Name the count of coral balls: this is $Z$

WHAT: We count only the coral (proton) balls and give that count a name: $Z$ .

WHY tag it with its own letter? Because this single number decides which element the atom is. One coral ball ⇒ hydrogen. Two ⇒ helium. Thirteen ⇒ aluminium. Change $Z$ and you have literally changed the substance. It deserves its own name because it is the atom's identity.

PICTURE: In the figure we've faded the mint balls to grey and lit up only the coral ones. Count the glowing balls — that count, written under the arrow, is $Z$ .

$Z = protons number of coral balls$

$Z$ — the label we chose (from German Zahl, "number").
"number of coral balls" — literally count the protons in the picture, nothing more.

Step 3 — Name the count of mint balls: this is $N$

WHAT: Now count only the mint (neutron) balls. Call that count $N$ .

WHY a separate letter? The mint balls add weight but no charge and no identity change. Two atoms can be the same element (same $Z$ ) yet hold different numbers of mint balls. To talk about that difference we need $N$ on its own.

PICTURE: This time we light up the mint balls and fade the coral. The glowing count under the arrow is $N$ .

$N = neutrons number of mint balls$

$N$ — our label for the neutron count.
These balls have no "+" or "−" mark, so they never touch the charge balance of Step 2.

Step 4 — Add the two counts: this is the mass number $A$

WHAT: Put the coral pile and the mint pile side by side and count all the balls in the bag together. That grand total is the mass number $A$ .

WHY does one total capture the mass? Because an electron is about $1836$ times lighter than a proton — so light it barely tips the scale. Almost all the weight sits in the nucleus, and every ball there (coral or mint) weighs almost exactly the same: one "unit." So "how heavy is the atom" ≈ "how many nucleons" ≈ the total ball count. That is why we can summarise mass with a plain integer.

PICTURE: The figure lines up the coral balls, then the mint balls, in a single row. Counting left-to-right across the whole row gives $A$ . The bracket underneath splits that same row into its coral part ( $Z$ ) and mint part ( $N$ ).

$A = coral count Z + mint count N$

$A$ — total nucleons = the full row length.
$Z$ — the coral segment of the row (from Step 2).
$N$ — the mint segment of the row (from Step 3).
" $+$ " — literally laying the two segments end to end.

Read term-by-term, $N = A - Z$ says: (all balls) minus (coral balls) = (mint balls). It cannot be otherwise, because coral and mint are the only two kinds.

Step 5 — Pack the counts onto one address: the nuclear symbol

WHAT: We stack $A$ (top-left) and $Z$ (bottom-left) onto the element letter X:

$_{Z}^{A} X$

WHY this layout? So a single glyph carries the full census. Top tells weight, bottom tells identity, and $N = A - Z$ falls out by subtraction. Every particle count is recoverable from these three characters.

PICTURE: The figure labels each slot with an arrow: top-left arrow → "total balls, $A$ "; bottom-left arrow → "coral balls, $Z$ "; and a dashed arrow shows the subtraction $A - Z$ pointing at the mint balls.

Reading $_{13}^{27} Al$ off the picture: coral $= 13$ , total $= 27$ , so mint $= 27 - 13 = 14$ , and (neutral) electrons $= 13$ .

Step 6 — Edge case: turning the atom into an ion

WHAT: An ion is an atom that has gained or lost some electrons. Its charge is written $q$ . We ask: what changes inside the bag?

WHY this is a "must-cover" case: beginners fear ions rearrange the nucleus. They do not. Ionization is an outside event — only the far-away slate dots (electrons) move. The coral and mint balls are untouched, so $Z$ , $N$ , and $A$ are all frozen.

PICTURE: Two panels. Left: neutral atom, electrons $= Z$ . Right: same bag (identical coral + mint), but with electrons added or removed. Only the orbiting dots changed count.

$electrons = protons, fixed Z - the charge q$

$Z$ — protons, still frozen in the nucleus.
$q$ — the written charge. A minus here means extra electrons.

Check both signs so no case surprises you:

Cation $X^{+}$ (lost 1 electron): $q = + 1$ , electrons $= Z - (+ 1) = Z - 1$ . Fewer dots. ✓
Anion $X^{-}$ (gained 1 electron): $q = - 1$ , electrons $= Z - (- 1) = Z + 1$ . More dots. ✓
For $_{8}^{16} O^{2 -}$ : $q = - 2$ , electrons $= 8 - (- 2) = 10$ . ✓

Step 7 — The three families, drawn as "which number is locked"

WHAT: Now we build isotopes, isobars, isotones by asking one question three ways: which of the trio $(Z, A, N)$ do the atoms share?

WHY draw them together? The names sound alike and get mixed up. But visually they are three different "locks." Lock $Z$ → same element, different weight. Lock $A$ → same weight, different element. Lock $N$ → same mint count. One picture kills the confusion.

PICTURE: Three rows of bags.

Isotopes row: the coral count is identical in every bag (same $Z$ ), only mint balls vary.
Isobars row: the total balls per bag is identical (same $A$ ), but the coral/mint split differs.
Isotones row: the mint count is identical (same $N$ ), coral differs.

The one-picture summary

Everything above compresses into a single flow: count coral → $Z$ , count mint → $N$ , add them → $A$ , stack onto X → the symbol, then lock one number → a family. The final figure lays this pipeline out with the three family-locks branching off the end.

$A = Z + N N = A - Z e^{-} = Z - q$

Recall Feynman retelling of the whole walkthrough

Picture a little bag with two kinds of ball — coral ones that carry a plus sign, and blank mint ones. Count the coral balls; that number is $Z$ , and it is the name of the atom (13 coral = aluminium, always). Count the mint balls; that's $N$ , extra weight with no name-changing power. Tip every ball onto a scale — since each weighs one unit and the far-off electrons weigh basically nothing, the total ball count is the atom's weight, which we call $A$ . So $A = Z + N$ , and if you want the mint balls back, just take the whole pile and remove the coral ones: $N = A - Z$ . Write it neatly as $_{Z}^{A} X$ — top is total, bottom is coral, and mint is the leftover. If someone adds or steals electrons (the tiny orbiting dots), the bag itself never changes — the coral and mint stay put — only the dot-count shifts, which is $e^{-} = Z - q$ . Finally, play a matching game with many bags: lock the coral count and you get isotopes (same element, different weight); lock the total and you get isobars (same weight, different element); lock the mint count and you get isotones. Same simple counting, three fancy Greek names.

Recall Quick self-test

Neutrons in $_{26}^{56} Fe$ ? ::: $56 - 26 = 30$ . Electrons in $_{8}^{16} O^{2 -}$ ? ::: $8 - (- 2) = 10$ . $_{1}^{3} H$ and $_{2}^{3} He$ — which family? ::: Isobars (same $A = 3$ , different $Z$ ). Which number never changes when an atom is ionized? ::: $Z$ (the coral/proton count).

Connections

Rutherford Model of the Atom — why the mass lives in the nucleus at all.
Discovery of Protons, Neutrons and Electrons — the coral, mint, and dot particles by name.
Average Atomic Mass and Isotopic Abundance — why weighing many isotopes gives non-integer masses.
Radioactivity and Nuclear Stability — how the $N / Z$ split decides survival.
Periodic Table and Atomic Number — ordering elements by $Z$ , the coral count.