Atomic Structure Formula Sheet — JEE Main Chemistry

Q: What are the most important Atomic Structure formulas for JEE Main?

This Atomic Structure formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Nature of electromagnetic radiation, Photoelectric effect, Spectrum of hydrogen atom, Bohr model of hydrogen atom, Postulates and derivations — each shown with the key result and, where useful, a worked example.

Q: Is this Atomic Structure formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Atomic Structure formulas?

Blurt the Atomic Structure formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Atomic Structure formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 21 sub-topics

Nature of electromagnetic radiation
Photoelectric effect
Spectrum of hydrogen atom
Bohr model of hydrogen atom
Postulates and derivations
Limitations of Bohr's model
Dual nature of matter
de Broglie's relationship
Heisenberg uncertainty principle
Elementary ideas of quantum mechanics
Quantum mechanical model of atom
Atomic orbitals as one-electron wave functions
Variation of ψ and ψ² with r
Quantum numbers and significance
Shapes of s, p and d-orbitals
Electron spin and spin quantum number
Aufbau principle
Pauli's exclusion principle
Hund's rule
Electronic configuration of elements
Extra stability of half-filled and completely filled orbitals

Subatomic Particles & Atomic Models

★ Remember · This page — not in the 2026 Chemistry syllabus

The 2026 Unit 2 starts at 'nature of electromagnetic radiation'; discovery of subatomic particles, Thomson/Rutherford models and isotopes/isobars/isotones are not in it (Rutherford's model is in PHYSICS Unit 18). Kept as background for older questions;

Z

A

and

N

are still needed everywhere.

Particle	Charge	Mass (u)
Electron $e^{-}$	$- 1.6 \times 1 0^{- 19}$ C	$0.00055$
Proton $p^{+}$	$+ 1.6 \times 1 0^{- 19}$ C	$1.00727$
Neutron $n^{0}$	$0$	$1.00867$

Atomic models

▸ $Thomson$ (plum-pudding): $+$ sphere with embedded electrons — explains neutrality, not stability.
▸ $Rutherford$ ( $α$ -scattering): tiny dense $+$ nucleus ( $\sim 1 0^{- 15}$ m), electrons orbit; atom mostly empty.

Atomic & mass number

Z = protons = electrons (neutral); A = Z + N;_{Z}^{A} X

N =

neutrons; nuclear charge

= + Z e

Rutherford nuclear atom with a central positive nucleus and orbiting electrons, plus the A-Z-X notation — Nucleus holds charge & mass; electrons orbit.

Comparative: same-something nuclei

Family	Same	Example
Isotopes	$Z$	$^{1}$ H, $^{2}$ H, $^{3}$ H
Isobars	$A$	$_{6}^{14}$ C, $_{7}^{14}$ N
Isotones	$N$	$_{6}^{14}$ C, $_{8}^{16}$ O

🚫 Examiner Trap · Particles & models

(1) Z (protons) defines the

element

;

A = Z + N

. (2) Isotopes (same Z) have the SAME chemistry; isobars (same A) are different elements. (3) Rutherford's flaw: an orbiting electron accelerates

\Rightarrow

must radiate & spiral in (

\sim 1 0^{- 8}

s) — can't explain stability/line spectra. (4) Neutral atom: electrons

= Z

, NOT

= A

EM Radiation, Planck & Photoelectric Effect

Wave description

c = ν λ; \overset{ν}{ˉ} = \frac{1}{λ} (cm^{- 1})

c = 3 \times 1 0^{8}

m/s;

\overset{ν}{ˉ} =

wavenumber (waves per cm)

Planck's quantum

E = h ν = \frac{h c}{λ}; h = 6.626 \times 1 0^{- 34} J s

energy absorbed/emitted only in discrete quanta (photons)

Einstein photoelectric

h ν = W_{0} + \frac{1}{2} m v^{2}, W_{0} = h ν_{0}

KE_{ma x} = h (ν - ν_{0})

;

W_{0} =

work function,

ν_{0} =

threshold frequency

Left: an electromagnetic wave with perpendicular E and B fields. Right: photoelectric kinetic energy rising linearly with frequency above the threshold — EM wave; photoelectrons only above $ν_{0}$ .

Photoelectric facts

▸Emission is $instantaneous$ and only if $ν \geq ν_{0}$ .
▸KE rises with $frequency$ , not intensity; $number$ of electrons rises with intensity.

🚫 Examiner Trap · EM & photoelectric

(1) Photoelectron KE depends on

frequency

, intensity only sets the count — a classic distractor. (2) No emission below

ν_{0}

at any intensity. (3) Wavenumber

\overset{ν}{ˉ} = 1/ λ

m^{- 1}

) is

not

frequency. (4) Energy of 1 mole of photons

= N_{A} h ν

, not

h ν

Bohr Model & Hydrogen Spectrum

Rydberg formula

\overset{ν}{ˉ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), R_{H} = 109677 cm^{- 1}

n_{2} > n_{1}

; emission line spectrum of atomic H

Series	$n_{1}$	Region
Lyman	1	UV
Balmer	2	Visible
Paschen	3	IR
Brackett	4	IR
Pfund	5	IR

Bohr's postulates

▸Electrons revolve in fixed $stationary orbits$ without radiating.
▸Angular momentum quantised: $mvr = \frac{nh}{2 π}$ .
▸Photon on a jump: $Δ E = h ν$ .

Radius & energy (H-like)

r_{n} = 52.9 \frac{n ^{2}}{Z} pm; E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} eV

one-electron species only: H, He

^{+}

, L

i^{2 +}

. Also

v_{n} = 2.18 \times 1 0^{6} \frac{Z}{n}

m/s and

E_{n} = - 2.18 \times 1 0^{- 18} \frac{Z ^{2}}{n ^{2}}

J; ionisation energy from

n = 1

+ 13.6 Z^{2}

eV. Bohr's condition is de Broglie's standing wave:

2 π r_{n} = nλ

Hydrogen energy levels with downward transitions producing the Lyman (UV), Balmer (visible) and Paschen (IR) series — Transitions to $n = 1, 2, 3$ give the spectral series.

🚫 Examiner Trap · Bohr & spectrum

(1) Bohr works ONLY for one-electron species (use

Z^{2}

for He

^{+}

). (2)

Balmer

alone is visible; Lyman is UV. (3) Number of spectral lines from level n to ground

= n (n - 1) /2

. (4) Bohr fails for multi-electron atoms, fine structure, Zeeman/Stark, and ignores the electron's wave nature.

Dual Nature & Uncertainty Principle

de Broglie

λ = \frac{h}{m v} = \frac{h}{p} = \frac{h}{2 m KE}

every moving particle has a matter wave; confirmed by electron diffraction

When significant?

▸ $Microscopic$ (electron): measurable $λ$ , wave effects matter.
▸ $Macroscopic$ : $λ$ undetectably small.

Heisenberg uncertainty

Δ x \cdot Δ p \geq \frac{h}{4 π}; Δ x \cdot Δ v \geq \frac{h}{4 πm}

position & momentum can't both be exact

Top: a matter wave illustrating de Broglie wavelength. Bottom: a localized wave packet illustrating position-momentum uncertainty — Wave nature of matter; position–momentum trade-off.

🚫 Examiner Trap · Dual nature & uncertainty

(1) Same KE:

λ \propto 1/ m

(heavier

\to

shorter); same speed:

λ \propto 1/ m

. (2) Uncertainty uses

h /4 π

, not h. (3) The principle forbids a definite

orbit

\Rightarrow

position becomes a

probability

(orbital). (4) For macroscopic bodies

λ

and uncertainty are negligible — they have definite paths.

Quantum Numbers

Orbital: A one-electron wave function

ψ

(from

\hat{H} ψ = E ψ

ψ

has no physical meaning, but

∣ ψ ∣^{2}

is the

probability density

— where the electron is most likely found.

Comparative: the four quantum numbers

QN	Values	Tells
$n$	$1, 2, 3 \dots$	shell, size, energy
$l$	$0$ to $n - 1$	subshell shape ( $s, p, d, f$ )
$m_{l}$	$- l$ to $+ l$	orientation ( $2 l + 1$ )
$m_{s}$	$\pm \frac{1}{2}$	spin

Counting

orbitals in subshell = 2 l + 1; orbitals in shell = n^{2}; max e^{-} = 2 n^{2}

s:1, p:3, d:5, f:7 orbitals (2 e

^{-}

each)

🚫 Examiner Trap · Quantum numbers

(1) l runs

0

n - 1

(so

n = 2

has s,p only — no

2 d

). (2)

Pauli

: no two electrons share all four QN

\Rightarrow

max 2/orbital, opposite spins. (3)

m_{l}

has

2 l + 1

values (orientations), not l. (4) Max electrons in shell

= 2 n^{2}

, in subshell

= 2 (2 l + 1)

Orbitals — Shapes, Nodes & Energy

Shapes

▸ $s$ : spherical (1, $l = 0$ ).
▸ $p$ : dumbbell, nodal plane through nucleus (3: $p_{x}, p_{y}, p_{z}$ , $l = 1$ ).
▸ $d$ : cloverleaf / double-dumbbell (5, $l = 2$ ).

Nodes

radial = n - l - 1, angular = l, total = n - 1

a node is where

∣ ψ ∣^{2} = 0

For hydrogen 1s and 2s: psi, psi squared and radial probability 4 pi r squared psi squared against r. 1s radial probability peaks at a0; 2s psi changes sign at the radial node at 2 a0 and its radial probability has a small inner peak and a larger outer peak — $ψ$ vs $ψ^{2}$ vs radial probability for 1s and 2s (syllabus item).

Reading

ψ

ψ^{2}

and

4 π r^{2} ψ^{2}

▸ $ψ^{2}$ (probability PER VOLUME) is largest AT the nucleus for every s orbital.
▸Radial probability $4 π r^{2} ψ^{2}$ is zero at the nucleus; for 1s it peaks at $r = a_{0} = 52.9$ pm (Bohr's radius).
▸2s: $ψ$ changes sign at $r = 2 a_{0}$ (one radial node, $n - l - 1 = 1$ ); the radial probability has a small inner and a larger outer peak, so 2s is bigger than 1s.

Comparative: orbital energy order

	H / one-electron	Multi-electron
Depends on	$n$ only	$n$ and $l$
$2 s$ vs $2 p$	equal (degenerate)	$2 s < 2 p$
Rule	—	lower $(n + l)$ first

(n + l) rule

lower (n + l) \Rightarrow lower energy; ties \to lower n

4 s (4) < 3 d (5)

, so

4 s

fills first

🚫 Examiner Trap · Orbitals & energy

(1) In

H

all subshells of a shell are degenerate (

2 s = 2 p

); in multi-electron atoms shielding splits them (

2 s < 2 p

). (2) Total nodes

= n - 1

, radial

= n - l - 1

— a

3 p

has

1

radial

+ 1

angular. (3) Penetration

s > p > d > f

\Rightarrow

s is lowest energy in a shell. (4)

∣ ψ ∣^{2}

, not

ψ

, is the probability.

Electronic Configuration

Three filling rules

▸ $Aufbau$ : fill by increasing energy — $1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p 5 s 4 d 5 p 6 s \dots$
▸ $Pauli$ : max 2 e $^{-}$ /orbital, opposite spins.
▸ $Hund$ : singly fill degenerate orbitals (parallel spins) before pairing.

Left: the diagonal (n+l) order of filling orbitals. Right: orbital-box diagrams for nitrogen and oxygen illustrating Hund's rule and Pauli's principle — Diagonal $(n + l)$ filling; Hund's parallel spins.

Writing configurations

▸Use the noble-gas core: Na $= [Ne] 3 s^{1}$ ; valence $=$ outermost shell.
▸ $4 s$ fills $before$ $3 d$ , but $4 s$ empties $first$ on ionisation.
▸Unpaired electrons n give the spin-only magnetic moment $μ = n (n + 2)$ BM (e.g. F $e^{3 +}$ $3 d^{5}$ : $n = 5$ , $μ = 5.92$ BM); $n = 0$ means diamagnetic.

🚫 Examiner Trap · Electronic configuration

(1) Cr

= [Ar] 3 d^{5} 4 s^{1}

and Cu

= [Ar] 3 d^{10} 4 s^{1}

(NOT

d^{4} s^{2}

d^{9} s^{2}

) — half/fully-filled d gain exchange-energy stability. (2) On ionisation electrons leave

4s before 3d

(Fe

^{2 +} = 3 d^{6}

, not

3 d^{4} 4 s^{2}

). (3) Hund: maximise parallel spins first. (4) Valence electrons are outermost-shell, not the last-filled subshell for transition metals.

More JEE Main Chemistry formula sheets

Frequently Asked Questions

What are the most important Atomic Structure formulas for JEE Main?

This Atomic Structure formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Nature of electromagnetic radiation, Photoelectric effect, Spectrum of hydrogen atom, Bohr model of hydrogen atom, Postulates and derivations — each shown with the key result and, where useful, a worked example.

Is this Atomic Structure formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Atomic Structure formulas?

Blurt the Atomic Structure formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheetsJEE Main previous-year papersmost important chapters.