Atomic number Z, mass number A, isotopes, isobars, isotones: Worked examples

This page is a drill room. The parent note built the tools; here we throw every kind of question at them so no exam variant catches you off guard. Before any numbers, we lay out a map of all possible cases — then hit each cell with a fully worked example.

The scenario matrix

Below is the complete list of "shapes" a question can take. Every worked example is tagged with the cell it covers, so you can see the coverage is total.

Cell	Case class	What makes it tricky	Example
A	Neutral atom, plain count	nothing — the baseline	Ex 1
B	Positive ion (cation)	electrons $< Z$	Ex 2
C	Negative ion (anion)	electrons $> Z$	Ex 2
D	Classify a pair — isotopes	same $Z$	Ex 3
E	Classify a pair — isobars	same $A$ , different $Z$	Ex 3
F	Classify a pair — isotones	same $N$ only	Ex 4
G	Degenerate input: $N = 0$	a nucleus with no neutrons	Ex 5
H	Degenerate input: $Z = 0$ ? / lightest limits	what is the smallest $Z$ possible	Ex 5
I	Reverse problem: given counts, build the symbol	work backwards	Ex 6
J	Real-world word problem	strip the story, keep the numbers	Ex 7
K	Exam twist: trap that looks like one family but is another	forecast-then-verify	Ex 8
L	Limiting / big-nucleus sanity check ( $N / Z$ trend)	why heavy atoms need extra neutrons	Ex 9

The figure below is the picture we will lean on the whole page: a nucleus is just a bag of two kinds of ball.

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

Look at it: burnt-orange balls are protons (count them → that's $Z$ ), teal balls are neutrons (count them → that's $N$ ), and all the balls together are the nucleons (count them → that's $A$ ). Every example is just recounting this bag.

The examples

Cell A — the baseline count

Worked example Example 1 — full particle count for

$_{15}^{31} P$ Forecast: Before reading on, guess: how many protons, neutrons, and electrons does neutral phosphorus-31 have? Write three numbers down.

Step 1 — read the trio off the symbol. $Z = 15$ (bottom-left), $A = 31$ (top-left). Why this step? The nuclear symbol $_{Z}^{A} X$ hands you $Z$ and $A$ directly; everything else is derived from these two.

Step 2 — protons $= Z = 15$ . Why? $Z$ is defined as the proton count. No arithmetic needed.

Step 3 — neutrons $= A - Z = 31 - 15 = 16$ . Why? $A$ counts all nucleons (protons + neutrons); remove the protons to leave the neutrons. This is the rearranged $N = A - Z$ .

Step 4 — electrons $= 15$ . Why? "Neutral" means total charge $0$ : the $15$ positive protons must be balanced by $15$ negative electrons.

Verify: protons + neutrons $= 15 + 16 = 31 = A$ ✓. Charge $= (+ 15) + (- 15) = 0$ ✓ (neutral). Answer: $15$ p, $16$ n, $15$ e⁻.

Cells B & C — ions (electrons move, nucleus frozen)

Worked example Example 2 — two ions at once:

$_{12}^{24} Mg^{2 +}$ and $_{16}^{32} S^{2 -}$ Forecast: An ion has a charge. Does the charge change the number of protons? The number of neutrons? Guess which one moves.

Step 1 — the cation $_{12}^{24} Mg^{2 +}$ : trio. $Z = 12$ , $A = 24$ , so $N = 24 - 12 = 12$ . Why? The nucleus is untouched by ionization, so we read $Z, A$ exactly as for a neutral atom.

Step 2 — electrons of the cation $= Z - q = 12 - (+ 2) = 10$ . Why this step? Charge $q = + 2$ means the atom lost $2$ electrons, so it now has fewer electrons than protons. The formula $e^{-} = Z - q$ encodes "start with $Z$ electrons, subtract the charge."

Step 3 — the anion $_{16}^{32} S^{2 -}$ : trio. $Z = 16$ , $A = 32$ , so $N = 32 - 16 = 16$ .

Step 4 — electrons of the anion $= Z - q = 16 - (- 2) = 18$ . Why? A charge of $- 2$ means $2$ extra electrons; subtracting a negative adds. So the anion has $18$ electrons, more than its $16$ protons.

Verify: Cation: protons $12$ , electrons $10$ ⇒ net charge $+ 2$ ✓. Anion: protons $16$ , electrons $18$ ⇒ net charge $- 2$ ✓. In both, protons and neutrons never budged — only electrons did.

Cells D & E — classify a pair (isotopes vs isobars)

Worked example Example 3 — classify: (a)

$_{10}^{20} Ne$ & $_{10}^{22} Ne$ ; (b) $_{18}^{40} Ar$ & $_{20}^{40} Ca$ Forecast: For each pair, guess the family (isotope / isobar / isotone) before computing.

Step 1 — build the trio for all four species.

Species	$Z$	$A$	$N = A - Z$
$_{10}^{20} Ne$	10	20	10
$_{10}^{22} Ne$	10	22	12
$_{18}^{40} Ar$	18	40	22
$_{20}^{40} Ca$	20	40	20

Why this step? Classification is just "which of $Z, A, N$ matches?" You cannot see the match until all three are written for every atom.

Step 2 — pair (a): compare. Same $Z = 10$ (both neon), different $A$ ( $20$ vs $22$ ). Why? Matching $Z$ = "same place in the periodic table" = iso-tope. ⇒ isotopes.

Step 3 — pair (b): compare. Different $Z$ ( $18$ vs $20$ ⇒ different elements), same $A = 40$ . Why? Matching $A$ = "same weight" = iso-bar. ⇒ isobars.

Verify: In (a) the $N$ values differ ( $10 \neq = 12$ ) — consistent with "same element, different mass." In (b) the $N$ values also differ ( $22 \neq = 20$ ), so they are not isotones — matching $A$ with different $Z$ forces different $N$ , exactly what isobar means. ✓

Cell F — the isotone (only $N$ matches)

Recall Which family matches which quantity?

Same $Z$ ::: isotopes Same $A$ ::: isobars Same $N$ ::: isotones

Cells G & H — degenerate inputs (the edge of the table)

Worked example Example 5 — the strange nuclei:

$_{1}^{1} H$ (protium) and "can $Z = 0$ ?" Forecast: Ordinary hydrogen $_{1}^{1} H$ — how many neutrons? And could an atom ever have $Z = 0$ ?

Step 1 — protium's neutron count: $N = A - Z = 1 - 1 = 0$ . Why this step? The formula must survive its most extreme input. Here $A = Z$ , so $N$ collapses to zero: a nucleus made of a single proton, no neutrons at all. This is the lightest real nucleus and it is perfectly valid.

Step 2 — electrons of neutral protium $= Z = 1$ . Why? Neutrality still holds: one proton, one electron.

Step 3 — the $Z = 0$ question. $Z$ counts protons, and $Z$ is the element's identity. $Z = 1$ is hydrogen; there is no element with $Z = 0$ — a bag with zero protons has no chemical identity (a lone neutron is a particle, not an element). So $Z = 1$ is the smallest atomic number. Why include this? To fix the lower boundary of the whole scheme: $Z \geq 1$ always, and $N$ can be $0$ but $Z$ cannot.

Verify: Protium: $Z + N = 1 + 0 = 1 = A$ ✓. The formula $N = A - Z$ gracefully gives $0$ , never a negative number, because a real nucleus always has $A \geq Z$ (you can't have more protons than nucleons). ✓

Cell I — reverse problem (build the symbol)

Cell J — real-world word problem

Worked example Example 7 — the smoke-detector isotope

Statement: A household smoke detector uses americium-241. Americium sits at atomic number $95$ . How many neutrons are in one nucleus of this isotope, and how many electrons does a neutral americium-241 atom have?

Forecast: Strip the story to numbers. What is $A$ ? What is $Z$ ? Guess $N$ .

Step 1 — extract the numbers from the words. "americium-241" ⇒ $A = 241$ (the number after the name is always the mass number). "atomic number $95$ " ⇒ $Z = 95$ . Why this step? Word problems hide the trio inside prose. The named-number after an element is always $A$ ; the phrase "atomic number" is always $Z$ .

Step 2 — neutrons $= A - Z = 241 - 95 = 146$ . Why? Same universal formula; the physics does not care that it came wrapped in a story.

Step 3 — electrons (neutral) $= Z = 95$ . Why? Neutral ⇒ electrons balance protons.

Verify: $Z + N = 95 + 146 = 241 = A$ ✓. Units/sanity: neutron count ( $146$ ) exceeds proton count ( $95$ ) — expected for a heavy nucleus (see Cell L below). ✓

Cell K — the exam twist (looks like one family, is another)

Worked example Example 8 — the classic trap:

$_{6}^{14} C$ and $_{7}^{14} N$ Forecast: They both say " $14$ ". So they must have the same neutrons... right? Commit to yes or no before Step 1.

Step 1 — trio for both. $_{6}^{14} C$ : $N = 14 - 6 = 8$ . $_{7}^{14} N$ : $N = 14 - 7 = 7$ . Why this step? The shared " $14$ " is the mass number $A$ , not the neutron number. The only way to catch the trap is to actually compute $N$ for each.

Step 2 — identify the true family. Same $A = 14$ , different $Z$ ( $6 \neq = 7$ ) ⇒ isobars, not isotones. Their neutron counts ( $8$ vs $7$ ) are different. Why? Because $A = Z + N$ : if $A$ is fixed and $Z$ goes up by $1$ , then $N$ must go down by $1$ . Same- $A$ forbids same- $N$ when the elements differ.

Verify: $N (C) = 8$ , $N (N) = 7$ , and indeed $8 \neq = 7$ , so "same mass ⇒ same neutrons" is false. The forecast "yes" is the trap; the answer is no — they are isobars. ✓

Cell L — the limiting / heavy-nucleus trend

Worked example Example 9 — why does

$N$ outgrow $Z$ in big atoms? Statement: Compare the neutron-to-proton ratio $N / Z$ for light $_{2}^{4} He$ , mid $_{26}^{56} Fe$ , and heavy $_{92}^{238} U$ .

Forecast: Guess whether $N / Z$ stays near $1$ or grows as atoms get heavier.

Step 1 — compute $N$ and the ratio for each.

Nucleus	$Z$	$A$	$N = A - Z$	$N / Z$
$_{2}^{4} He$	2	4	2	$1.00$
$_{26}^{56} Fe$	26	56	30	$\approx 1.15$
$_{92}^{238} U$	92	238	146	$\approx 1.59$

Why this step? A "trend" only appears when you line up several cases and read the pattern in one column.

Step 2 — read the pattern. $N / Z$ climbs from $1.00 \to 1.15 \to 1.59$ : heavier nuclei carry proportionally more neutrons. Why does nature do this? Protons all repel each other electrically; the more protons you pack in, the more extra (chargeless) neutrons you need to space them out and hold the nucleus together. This is the seed of Radioactivity and Nuclear Stability — the $N / Z$ ratio decides which nuclei survive.

Verify: $2/2 = 1.00$ ✓; $30/26 = 1.1538 \dots \approx 1.15$ ✓; $146/92 = 1.5869 \dots \approx 1.59$ ✓, and the sequence is strictly increasing. ✓ (See Average Atomic Mass and Isotopic Abundance for how these real isotopes average out to the masses on the periodic table.)

Coverage check

Recall Did we hit every cell of the matrix?

A (baseline) → Ex 1 ::: ✓ B & C (cation, anion) → Ex 2 ::: ✓ D & E (isotopes, isobars) → Ex 3 ::: ✓ F (isotones) → Ex 4 ::: ✓ G & H ( $N = 0$ , smallest $Z$ ) → Ex 5 ::: ✓ I (reverse: build symbol) → Ex 6 ::: ✓ J (word problem) → Ex 7 ::: ✓ K (exam trap) → Ex 8 ::: ✓ L (heavy-nucleus limit) → Ex 9 ::: ✓

Connections

Yeh note Hinglish mein padho →
Discovery of Protons, Neutrons and Electrons — the particles every count above refers to.
Periodic Table and Atomic Number — used in Ex 6 to name the element from $Z$ .
Radioactivity and Nuclear Stability — the $N / Z$ trend of Ex 9.
Average Atomic Mass and Isotopic Abundance — where isotope counts turn into real masses.
Rutherford Model of the Atom — why the nucleus is the thing we are counting.