How many neutrons and protons are there in the following nuclei? $^{13}_{6}C,\ ^{16}_{8}O,\ ^{24}_{12}Mg,\ ^{56}_{26}Fe,\ ^{88}_{38}Sr$

How many neutrons and protons are there in the following nuclei? ¹³₆C,¹⁶₈O,²⁴₁₂Mg,⁵⁶₂₆Fe,⁸⁸₃₈Sr

Exercises · 2.3

^{13}_{6}C,\ ^{16}_{8}O,\ ^{24}_{12}Mg,\ ^{56}_{26}Fe,\ ^{88}_{38}Sr

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Concept understanding — Atomic Notation

Atomic Notation: The ID Card of an Atom

Imagine you're at a huge stadium filled with people. To identify any one person, you'd need more than just a name — you'd need their jersey number, their team, maybe their position. An atom is the same way. Just saying "carbon" or "oxygen" tells you the kind of atom, but not the full story. Atomic notation is the shorthand that gives you the complete identity.

The Intuition: Three Numbers, One Symbol

Every atom is built from three key particles:

Protons (positive charge) — these define what element it is.
Neutrons (no charge) — these add mass and stability.
Electrons (negative charge) — these orbit the nucleus and determine chemical behaviour.

Atomic notation packs all this information around the element's symbol. Think of it as a label with two numbers attached:

^{A}_{Z}\text{X}

Where:

X is the element symbol (e.g., C for carbon, O for oxygen).
Z is the atomic number (number of protons).
A is the mass number (protons + neutrons).

Note

The atomic number $Z$ is what makes an atom that element. Change $Z$ , and you change the element entirely. Carbon always has $Z=6$ ; if you change it to $7$ , it's nitrogen.

The Precise Statement

For any atom represented as $^{A}_{Z}\text{X}$ :

Atomic number $Z$ = number of protons = number of electrons (in a neutral atom).
Mass number $A$ = number of protons + number of neutrons.
Number of neutrons $N = A - Z$ .

A neutral atom has equal protons and electrons. If the atom gains or loses electrons, it becomes an ion — but the notation for the nucleus stays the same.

Example: Carbon-12

The most common carbon atom is written as:

^{12}_{6}\text{C}

From this:

$Z=6$ → 6 protons, and in a neutral atom, 6 electrons.
$A=12$ → total nucleons (protons + neutrons) = 12.
Neutrons $N = 12 - 6 = 6$ .

Why Two Numbers? Why Not Just One?

If you only knew the mass number $A$ , you wouldn't know the element — because different elements can have the same mass number (e.g., $^{14}\text{C}$ and $^{14}\text{N}$ both have $A=14$ ). If you only knew the atomic number $Z$ , you wouldn't know how many neutrons are present — and that matters for isotopes.

Important

Isotopes are atoms of the same element (same $Z$ ) with different numbers of neutrons (different $A$ ). For example:

$^{12}_{6}\text{C}$ (6 neutrons)
$^{13}_{6}\text{C}$ (7 neutrons)
$^{14}_{6}\text{C}$ (8 neutrons) All are carbon, but they have different masses and some are radioactive.

A Quick Table for Clarity

Notation	Element	Protons ( $Z$ )	Neutrons ( $N$ )	Mass Number ( $A$ )
$^{1}_{1}\text{H}$	Hydrogen	1	0	1
$^{4}_{2}\text{He}$	Helium	2	2	4
$^{23}_{11}\text{Na}$	Sodium	11	12	23
$^{235}_{92}\text{U}$	Uranium	92	143	235

Common Mistakes to Avoid

Watch out

Don't confuse $A$ and $Z$ . $Z$ is always the smaller number (protons), $A$ is the larger (total nucleons).
Don't assume $A$ equals atomic mass. Atomic mass is an average of all isotopes, not the mass number of a single atom.
Don't forget that $Z$ defines the element. If you see $^{14}_{7}\text{N}$ , that's nitrogen — not carbon, even though $A=14$ is the same as carbon-14.

The Bottom Line

Atomic notation $^{A}_{Z}\text{X}$ is the atom's ID card. It tells you:

What element (via $Z$ )
Which isotope (via $A$ )
How many neutrons (via $A-Z$ )

Once you can read this notation, you can decode the entire nuclear structure of any atom in a single glance.

Atomic notation, writing an element with its mass number and atomic number, is introduced early in the NCERT Class 11 Chemistry chapter on Structure of Atom, and "atomic notation definition, formula and examples" is a common search among students building their fundamentals for CBSE boards, JEE Main, and NEET. This notation is also used constantly in the Class 12 Physics Nuclei chapter, making it one of the most cross-referenced basics in PCM/PCB exam preparation.

For any nucleus written as $^{A}_{Z}X$ , the atomic number $Z$ gives the number of protons, and the neutron count is $A - Z$ . Applying this to the given nuclei: $^{13}_{6}C$ has 6 protons and 7 neutrons; $^{16}_{8}O$ has 8 protons and 8 neutrons; $^{24}_{12}Mg$ has 12 protons and 12 neutrons; $^{56}_{26}Fe$ has 26 protons and 30 neutrons; $^{88}_{38}Sr$ has 38 protons and 50 neutrons.

The idea: reading a nuclear symbol

Every nucleus is written in the standard notation $^{A}_{Z}X$ , where:

$X$ is the chemical symbol of the element.
$Z$ is the atomic number — the number of protons. This defines which element it is.
$A$ is the mass number — the total number of protons plus neutrons.

The number of neutrons $N$ is simply the difference: $N = A - Z$ .

That’s the entire principle. There is no hidden complexity — it’s just careful subtraction once you’ve correctly identified $Z$ and $A$ from the superscript and subscript.

Watch out

A common mistake is to confuse the mass number $A$ with the atomic mass (which is not an integer). Here, $A$ is always a whole number — it counts nucleons, not grams. Also, never assume the number of neutrons equals the number of protons; that’s only true for a few light stable nuclei.

Step-by-step for each nucleus

1. $^{13}_{6}C$

The subscript is $Z = 6$ , so there are 6 protons.

The superscript is $A = 13$ , so neutrons = $13 - 6 = 7$ .

2. $^{16}_{8}O$

$Z = 8$ → 8 protons.

$A = 16$ → neutrons = $16 - 8 = 8$ .

3. $^{24}_{12}Mg$

$Z = 12$ → 12 protons.

$A = 24$ → neutrons = $24 - 12 = 12$ .

4. $^{56}_{26}Fe$

$Z = 26$ → 26 protons.

$A = 56$ → neutrons = $56 - 26 = 30$ .

5. $^{88}_{38}Sr$

$Z = 38$ → 38 protons.

$A = 88$ → neutrons = $88 - 38 = 50$ .

Tip

Notice that for $^{16}_{8}O$ and $^{24}_{12}Mg$ , the number of protons equals the number of neutrons. This is a special case — these are “doubly magic” or light stable nuclei where $N = Z$ . For heavier elements like iron and strontium, neutrons outnumber protons, which is the general trend for stability.

✓Final answer

^{13}_{6}C

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$Q. The total number of electrons, protons and neutrons present in the three isotopes of hydrogen is (A) 3 (B) 5 (C) 6 (D) 9$
›Reveal solutionSolution
Add up protons + electrons + neutrons for protium, deuterium, and tritium separately, then sum all three totals to get 9.

Concept and Intuition

All three isotopes of hydrogen have the same atomic number (1 proton, 1 electron in the neutral atom) but differ in neutron number: protium has none, deuterium has one, and tritium has two. Summing every subatomic particle across all three isotopes is a simple counting exercise once the neutron counts are known.

Step-by-Step Solution

Protium ( $_1^1H$ ): 1 proton + 1 electron + 0 neutrons = 2.

Deuterium ( $_1^2H$ ): 1 proton + 1 electron + 1 neutron = 3.

Tritium ( $_1^3H$ ): 1 proton + 1 electron + 2 neutrons = 4.

Total = $2 + 3 + 4 = 9$ .

Common Mistakes

Forgetting protium has zero neutrons (mass number 1 = 1 proton + 0 neutrons).

Omitting the electron count (some students count only protons + neutrons, i.e. mass numbers, which would wrongly give $1+2+3=6$ ).

✓Final answer
$The correct option is (D) — 9.$

ANSWER: D
AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$^{23}\mathrm{Na^+}$
›Reveal solutionSolution
11 protons fixes the element as sodium; 12 neutrons gives mass number 23; 10 electrons (one fewer than protons) means it's the +1 cation, Na⁺.

Concept and Intuition

The number of protons defines the element (atomic number $Z$ ). The mass number $A=Z+$ neutrons. If the number of electrons differs from the number of protons, the species is an ion, with charge $=(\text{protons})-(\text{electrons})$ .

Step-by-Step Solution

Protons $=11 \Rightarrow$ element is sodium (Na, $Z=11$ ).

Mass number $A=$ protons + neutrons $=11+12=23$ .

Electrons $=10$ , protons $=11$ : charge $=+1$ (11 positive charges, 10 negative charges).

So the species is $^{23}\mathrm{Na^+}$ .

Common Mistakes

Picking neutral $^{23}$ Na (option B), forgetting to account for the electron deficit indicating a +1 ion.

Picking Neon (option D), confusing electron count (which matches a neon atom's 10 electrons) with the actual element identity, which is fixed by the proton count (11 = sodium), not electron count.

✓Final answer
$^{23}\mathrm{Na^+}$

ANSWER: A
AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$Q. Identify the group number of the element X which has 1.44% more number of neutrons than the protons. (Mass number of X is 127) (A) group 15 (B) group 16 (C) group 17 (D) group 14$
›Reveal solutionSolution
Solving $Z+1.44Z=127$ for the proton count gives $Z\approx52$ (with $N=75$ , matching the stated neutron-proton relationship almost exactly), placing element X in group 16.

Concept and Intuition

For a nuclide of known mass number $A=N+Z$ , if the neutron count is a known multiple/percentage-relation of the proton count, the two can be combined into a single equation in one unknown ( $Z$ ), pinning down the element uniquely from the periodic table, and hence its period and group.

Step-by-Step Solution

Let $Z$ = number of protons, $N$ = number of neutrons, with $N \approx 1.44Z$ (neutrons in a 1.44 : 1 relation to protons) and $N+Z=127$ .

Substitute: $Z + 1.44Z = 127 \Rightarrow 2.44Z = 127 \Rightarrow Z \approx 52.05$ , so $Z=52$ .

Check: $N = 127-52 = 75$ , and $75/52 = 1.442$ , consistent with the given relation.

Atomic number 52 corresponds to tellurium (Te), a p-block element.

Tellurium lies in period 5 and group 16 (the oxygen family: O, S, Se, Te, Po).

Common Mistakes

Solving for $Z$ but then misreading the periodic table position (confusing group 16 with group 15 or 17, the neighbouring p-block groups).

Rounding $Z$ incorrectly and landing on a nearby element instead of checking the neutron count consistency.

✓Final answer
$The correct option is (B) — group 16.$

ANSWER: B
AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
$Q. Which of the following statement is correct about the isotopes of hydrogen? (A) Internuclear distance in hydrogen and deuterium is same (B) Enthalpy of bond dissociation of deuterium is less than hydrogen. (C) Order of relative abundance of isotopes is H > T > D (D) Ratio of neutrons to protons is greater for D than T.$
›Reveal solutionSolution
Bond length is set by electron distribution, not nuclear mass, so H₂ and D₂ have essentially the same internuclear distance — making (A) the only correct statement.

Concept and Intuition

Isotopes have identical electron configurations and hence nearly identical chemical/bonding behavior (bond lengths depend on electronic structure). However, the heavier isotope's nucleus has lower zero-point vibrational energy (due to larger reduced mass), making its bond dissociation enthalpy slightly higher, not lower — a classic isotope effect that trips up statement (B).

Step-by-Step Solution

(A): Bond length in a diatomic molecule depends on the electron cloud/orbital overlap, which is essentially unchanged by isotopic substitution (protons, deuterons and tritons have the same charge and nearly the same electronic environment). So $r(H_2)\approx r(D_2)$ . TRUE.

(B): Because $D$ is heavier, its zero-point energy is lower, meaning MORE energy is needed to dissociate it — so BDE(D₂) > BDE(H₂), making the statement ("enthalpy of D is less than H") FALSE.

(C): Natural abundance is H (99.98%) ≫ D (0.02%) ≫ T (trace, radioactive), so the correct order is H > D > T, not H > T > D. FALSE.

(D): Neutron-to-proton ratio: D has 1 neutron/1 proton = 1; T has 2 neutrons/1 proton = 2. T's ratio is larger, so "greater for D than T" is FALSE.

Only (A) is correct.

Common Mistakes

Assuming heavier isotopes have weaker bonds by analogy with reduced vibrational frequency, without realizing zero-point energy considerations reverse this for BDE.

Overlooking that tritium, despite being "heavier" in name recognition, is actually the rarest of the three isotopes.

✓Final answer
$The correct option is (A) — Internuclear distance in hydrogen and deuterium is same.$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$^{12}\mathrm{C}$
›Reveal solutionSolution
Natural carbon's two stable isotopes are $^{12}\mathrm{C}$ and $^{13}\mathrm{C}$ ; $^{14}\mathrm{C}$ , though naturally occurring in trace amounts, is radioactive, not stable.

Concept and Intuition

Isotopes of an element share the same atomic number but differ in mass number (neutron count). Naturally occurring carbon is a mixture of isotopes, but only two of them are stable indefinitely — the rest, like $^{14}\mathrm{C}$ , undergo radioactive decay (in $^{14}\mathrm{C}$ 's case, this decay is the basis of radiocarbon dating).

Step-by-Step Solution

Carbon's naturally occurring isotopes include $^{12}\mathrm{C}$ , $^{13}\mathrm{C}$ , and trace $^{14}\mathrm{C}$ .

$^{12}\mathrm{C}$ (about 98.9% abundance) and $^{13}\mathrm{C}$ (about 1.1% abundance) are both stable (non-radioactive).

$^{14}\mathrm{C}$ is present only in trace amounts and is radioactive, decaying via beta emission with a half-life of about 5730 years — it is not classified as "stable."

Therefore the two stable isotopes are $^{12}\mathrm{C}$ and $^{13}\mathrm{C}$ , matching option (B).

Common Mistakes

Including $^{14}\mathrm{C}$ among the "stable" isotopes simply because it occurs naturally (natural occurrence does not imply stability).

Selecting nonexistent or irrelevant mass numbers like $^{11}\mathrm{C}$ or $^{15}\mathrm{C}$ , which are not significant naturally occurring carbon isotopes.

✓Final answer
$^{12}\mathrm{C}$

ANSWER: B
AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
$+3$
›Reveal solutionSolution
$X^{3+} = [Ar]3d^6$ has 24 electrons, so the atomic number of $X$ is $27$ .

Concept and Intuition

The electronic configuration of an ion tells us its electron count directly. A $+3$ charge means three electrons were lost relative to the neutral atom, so the atomic number equals the ion's electron count plus 3.

Step-by-Step Solution

Argon core $[Ar]$ has $18$ electrons.

$[Ar]3d^6$ therefore has $18 + 6 = 24$ electrons — this is $X^{3+}$ .

Since $X^{3+}$ lost 3 electrons, neutral $X$ has $24 + 3 = 27$ electrons.

Atomic number $Z = 27$ , which is cobalt (Co).

Common Mistakes

Forgetting to add back the 3 lost electrons and reporting $24$ (option B is $24$ ).

Miscounting the argon core as anything other than 18 electrons.

✓Final answer
$27$

ANSWER: C

Note

This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.
AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$^{13}_{6}C$
›Reveal solutionSolution
This tests basic nuclear notation $^A_ZX$ : protons $=Z$ , neutrons $=A-Z$ , electrons $=Z$ for a neutral atom — giving 6, 7, 6 for $^{13}_6C$ — answer (A).

Concept and Intuition

In the notation $^A_Z X$ , the subscript $Z$ (atomic number) is the number of protons, which also equals the number of electrons in a neutral atom, and the superscript $A$ (mass number) is the total number of nucleons (protons + neutrons). So neutrons are simply obtained by subtracting protons from the mass number.

Step-by-Step Solution

From $^{13}_6C$ : $Z=6$ (atomic number), $A=13$ (mass number).

Protons $= Z = 6$ .

Electrons (neutral atom) $= Z = 6$ .

Neutrons $= A - Z = 13 - 6 = 7$ .

So (protons, neutrons, electrons) $=(6,7,6)$ .

Common Mistakes

Swapping mass number and atomic number, e.g. thinking there are 13 protons.

Forgetting neutrons = mass number minus atomic number, not mass number itself.

✓Final answer
$The correct option is (A) — 6, 7, 6.$

ANSWER: A

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Q.How many neutrons and protons are there in the following nuclei? 613C, 816O, 1224Mg, 2656Fe, 3888Sr^{13}_{6}C,\ ^{16}_{8}O,\ ^{24}_{12}Mg,\ ^{56}_{26}Fe,\ ^{88}_{38}Sr613​C, 816​O, 1224​Mg, 2656​Fe, 3888​Sr

The idea: reading a nuclear symbol

Step-by-step for each nucleus

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Q.How many neutrons and protons are there in the following nuclei? $^{13}_{6}C,\ ^{16}_{8}O,\ ^{24}_{12}Mg,\ ^{56}_{26}Fe,\ ^{88}_{38}Sr$

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