An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Exercises · 2.42

Q. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

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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

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Solving $n = 1.317\,p$ together with $p + n = 81$ gives $p = 35$ , so the element is bromine. The symbol is $^{81}_{35}\text{Br}$ .

Let $p$ = number of protons and $n$ = number of neutrons.

Neutron–proton relation ("31.7% more neutrons than protons"):

n = p + 0.317\,p = 1.317\,p

Mass-number constraint:

p + n = 81

Substitute and solve:

$p + 1.317\,p = 81 \quad\Rightarrow\quad 2.317\,p = 81 \quad\Rightarrow\quad p = \frac{81}{2.317} \approx 35$ …

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

KCET 2025Set D-41 markMCQ
$Q. The number of orbitals associated with ‘N’ shell of an atom is (A) 16 (B) 32 (C) 3 (D) 4$
›Reveal solutionSolution

N shell $\Rightarrow n = 4$ ; the number of orbitals in a shell is $n^2 = 16$ (the number of electrons would be $2n^2 = 32$ , which is the trap).

Step 1 — Identify the shell.

The spectroscopic shell letters map to principal quantum numbers in order:
$K \to n=1,\quad L \to n=2,\quad M \to n=3,\quad \mathbf{N \to n=4}$
Step 2 — List the subshells allowed for $n = 4$ .

The azimuthal quantum number obeys $l = 0, 1, \dots, (n-1)$ , so for $n=4$ : $l = 0, 1, 2, 3$ — that is $4s, 4p, 4d, 4f$ .

Step 3 — Count the orbitals in each subshell.

For a given $l$ the magnetic quantum number takes $m_l = -l \dots 0 \dots +l$ , which is $(2l+1)$ values — and each value of $m_l$ is one orbital:

Subshell $l$ Orbitals $(2l+1)$
$4s$ 0 1
$4p$ 1 3
$4d$ 2 5
$4f$ 3 7

Step 4 — Add them up.
$1 + 3 + 5 + 7 = 16$
This reproduces the general result …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. An Anion formed by an atom carries one unit of negative charge and has a mass number equal to 37u. If this Anion contains 11.1% more neutrons than electrons, what is the correct number of protons, neutrons and electrons in it? (A) protons=17 electrons=18 neutrons=20 (B) protons=16 electrons=16 neutrons=21 (C) protons=19 electrons=19 neutrons=18 (D) protons=18 electrons=18 neutrons=19$
›Reveal solutionSolution

Solving the neutron-excess condition for a singly-charged anion of mass 37 gives 17 protons, 18 electrons, 20 neutrons.

Set up (anion charge $-1$ $\Rightarrow$ electrons $= p+1$ ; mass number $37 \Rightarrow$ neutrons $= 37-p$ ):

11.1% more neutrons than electrons:
$37-p = 1.111\,(p+1)$
$37 - 1.111 = 1.111p + p \Rightarrow 35.889 = 2.111p \Rightarrow p = 17$ …
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Subshell	$l$	Orbitals $(2l+1)$
$4s$	0	1
$4p$	1	3
$4d$	2	5
$4f$	3	7

←Previous question · exercises2.41 — Symbols ⁷⁹₃₅Br and ⁷⁹Br can be written, whereas symbols ³⁵₇₉Br and ³⁵B…→Next question · exercises2.43 — An ion with mass number 37 possesses one unit of negative charge. If t…

Q.An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

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