Chemistry Atomic Structure

Atomic Structure Formula Sheet

All key Atomic Structure formulas: photon energy, Bohr model, Rydberg, de Broglie, uncertainty, quantum numbers & nodes for JEE Main & Advanced quick revision.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula, constant, and high-yield fact from the Atomic Structure chapter, grouped by sub-topic for last-minute revision.

Key Constants

ConstantSymbolValue
Planck’s constant$h$$6.626 \times 10^{-34}$ J·s
Speed of light$c$$3 \times 10^8$ m/s
Rydberg constant$R_H$$1.097 \times 10^7$ m⁻¹
First Bohr radius$a_0$$0.529$ Å
Coulomb constant$k = \frac{1}{4\pi\epsilon_0}$$9 \times 10^9$ N·m²/C²
Electron mass$m_e$$9.1 \times 10^{-31}$ kg
eV to Joule$1\text{ eV} = 1.6 \times 10^{-19}$ J
Unit shortcuts

1 nm = 10⁻⁹ m  |  1 Å = 10⁻¹⁰ m  |  1 pm = 10⁻¹² m. Visible light: 400 nm (violet) → 700 nm (red).


Electromagnetic Radiation

QuantityFormulaNotes
Wave relation$c = \nu\lambda$$c = 3\times10^8$ m/s in vacuum
Wave number$\bar{\nu} = \dfrac{1}{\lambda}$unit m⁻¹
Photon energy$E = h\nu = \dfrac{hc}{\lambda}$higher $\nu$ → higher $E$
Photon momentum$p = \dfrac{h}{\lambda} = \dfrac{E}{c}$photons are massless but carry momentum
$$\boxed{E = h\nu = \frac{hc}{\lambda}}$$
Energy vs intensity
Photon energy depends on frequency ($E = h\nu$); intensity depends on the number of photons per second. Do not mix these up.

Photoelectric Effect

$$\boxed{h\nu = \phi + \tfrac{1}{2}mv_{max}^2 = h\nu_0 + KE_{max}}$$
QuantityFormulaNotes
Work function$\phi = h\nu_0$minimum energy to eject electron
Maximum KE$KE_{max} = h\nu - \phi$depends only on frequency
Stopping potential$eV_0 = KE_{max} = h\nu - \phi$$V_0 = \dfrac{h(\nu - \nu_0)}{e}$
Threshold wavelength$\lambda_0 = \dfrac{hc}{\phi}$maximum $\lambda$ that still ejects
Maximum velocity$v_{max} = \sqrt{\dfrac{2(h\nu - \phi)}{m}}$

$V_0$ vs $\nu$ graph: slope $= \dfrac{h}{e}$, x-intercept $= \nu_0$, y-intercept $= -\dfrac{\phi}{e}$.

High-yield facts

Below $\nu_0$: no emission at any intensity. One photon ejects at most one electron. $KE_{max}$ and $V_0$ are independent of intensity. Threshold frequency is a minimum; threshold wavelength is a maximum.


Bohr’s Model (hydrogen-like, nuclear charge $Ze$)

Postulates: quantized circular orbits (no radiation in a stationary state), quantized angular momentum, and energy emitted/absorbed only on transitions.

$$\boxed{mvr = \frac{nh}{2\pi} = n\hbar} \qquad \boxed{\Delta E = E_2 - E_1 = h\nu}$$
QuantityFormulaScaling
Force balance$\dfrac{mv^2}{r} = \dfrac{kZe^2}{r^2}$
Radius of nth orbit$r_n = \dfrac{0.529\,n^2}{Z}$ Å$r_n \propto \dfrac{n^2}{Z}$
Velocity in nth orbit$v_n = \dfrac{2.18\times10^6\,Z}{n}$ m/s$v_n \propto \dfrac{Z}{n}$
Energy of nth orbit$E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV$= -\dfrac{2.18\times10^{-18}Z^2}{n^2}$ J
Energy ratio$\dfrac{E_{n_1}}{E_{n_2}} = \dfrac{n_2^2}{n_1^2}$
Orbital period$T_n = \dfrac{2\pi r_n}{v_n}$$\propto \dfrac{n^3}{Z^2}$
Frequency of revolution$f_n = \dfrac{v_n}{2\pi r_n}$$\propto \dfrac{Z^2}{n^3}$
Ionization energy$IE = 0 - E_1 = 13.6\,Z^2$ eVfrom ground state to $\infty$
$$\boxed{r_n = \frac{0.529\,n^2}{Z}\ \text{Å} \qquad E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV}}$$

Hydrogen energy levels: $E_1 = -13.6$, $E_2 = -3.4$, $E_3 = -1.51$, $E_4 = -0.85$ eV; $E_\infty = 0$.

Ionization energies: H (Z=1) → 13.6 eV; He⁺ (Z=2) → 54.4 eV; Li²⁺ (Z=3) → 122.4 eV.

Magic numbers + energy identity

Memorize 0.529 Å, 2.18 × 10⁶ m/s, 13.6 eV. Also: Total energy $= -\,KE = \tfrac{1}{2}\,PE$, i.e. $E_n = -KE$.


Hydrogen Spectrum

$$\boxed{\frac{1}{\lambda} = \bar{\nu} = R_H Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)}$$

with $n_2 > n_1$ (final state $n_1$). Equivalently $h\nu = -13.6 Z^2\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right)$ eV.

Series$n_1$Region$\lambda_{limit}$ (nm)
Lyman1UV91.2
Balmer2Visible364.6
Paschen3Infrared820.4
Brackett4Far IR
Pfund5Far IR
QuantityFormulaNotes
Series limit$\dfrac{1}{\lambda_{limit}} = R_H Z^2 \dfrac{1}{n_1^2}$$n_2 \to \infty$ (shortest $\lambda$)
Lines from level $n$$N = \dfrac{n(n-1)}{2}$electron de-exciting from $n$ to ground
Lines for $n_2 \to n_1$$N = \dfrac{(n_2-n_1)(n_2-n_1+1)}{2}$
Spectrum mnemonics

Lyman → lowest level (n=1) → UV; Balmer → n=2 → Visible (only series in visible); Paschen/Brackett/Pfund → IR. First line of a series = longest $\lambda$; series limit = shortest $\lambda$. For He⁺ multiply by $Z^2 = 4$.


Dual Nature of Matter

$$\boxed{\lambda = \frac{h}{mv} = \frac{h}{p}}$$
QuantityFormulaNotes
de Broglie (KE form)$\lambda = \dfrac{h}{\sqrt{2m\,KE}}$same KE → $\lambda \propto \dfrac{1}{\sqrt{m}}$
Accelerated charge $q$ through $V$$\lambda = \dfrac{h}{\sqrt{2mqV}}$$KE = qV$
Electron through $V$ (volts)$\lambda = \dfrac{12.27}{\sqrt{V}}$ Åquick-calc shortcut
Thermal wavelength$\lambda = \dfrac{h}{\sqrt{3mkT}}$particles at temperature $T$
Bohr quantization$2\pi r = n\lambda$gives $mvr = \dfrac{nh}{2\pi}$

Heisenberg Uncertainty Principle

$$\boxed{\Delta x \cdot \Delta p \geq \frac{h}{4\pi}}$$

Equivalent forms: $\Delta x \cdot m\Delta v \geq \dfrac{h}{4\pi}$ and $\Delta E \cdot \Delta t \geq \dfrac{h}{4\pi}$.

Remember

$\lambda \propto \tfrac{1}{m}$ and $\lambda \propto \tfrac{1}{v}$; same KE → $\lambda \propto \tfrac{1}{\sqrt{m}}$; same momentum → same $\lambda$. Conjugate pairs: $(x, p)$, $(E, t)$, $(\theta, L)$. Uncertainty is a fundamental property of nature, not a measurement limit.


Quantum Mechanical Model

ItemExpressionNotes
Schrödinger equation$\hat{H}\psi = E\psi$$\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}+\dfrac{8\pi^2 m}{h^2}(E-V)\psi = 0$
Probability density$\lvert\psi\rvert^2$$\psi$ must be single-valued, continuous, finite
Normalization$\int \lvert\psi\rvert^2\,dV = 1$total probability = 1
Orbital angular momentum$L = \sqrt{l(l+1)}\cdot\dfrac{h}{2\pi}$depends on $l$
Spin angular momentum$S = \sqrt{s(s+1)}\cdot\dfrac{h}{2\pi} = \sqrt{\tfrac{3}{4}}\cdot\dfrac{h}{2\pi}$$s = \tfrac{1}{2}$

Four Quantum Numbers

QNSymbolAllowed valuesDetermines
Principal$n$$1, 2, 3, \dots$shell, size, energy
Azimuthal$l$$0$ to $(n-1)$subshell, shape
Magnetic$m_l$$-l$ to $+l$ ($2l+1$ values)orbital orientation
Spin$m_s$$+\tfrac{1}{2}$ or $-\tfrac{1}{2}$electron spin

Counts: orbitals in a subshell $= 2l+1$; orbitals in a shell $= n^2$; max electrons in subshell $= 2(2l+1)$; max electrons in shell $= 2n^2$.

Validity checks

$l < n$ always; $m_l$ ranges $-l \dots +l$. No 1p, 2d, 3f. For H-like atoms energy depends only on $n$ ($3s=3p=3d$); in multi-electron atoms it depends on $n$ and $l$.


Shapes of Orbitals & Nodes

Node typeFormulaDescription
Radial nodes$n - l - 1$spherical surfaces
Angular nodes$l$planes or cones
Total nodes$n - 1$sum of both
QuantityFormulaNotes
Radial probability$P(r) = 4\pi r^2 \lvert\psi\rvert^2$1s peak at $r = a_0 = 0.529$ Å
Subshell$l$ShapeAngular nodes$m_l$ values
s0Sphere00
p1Dumbbell (starts n=2)1 plane$-1, 0, +1$
d24-lobed / ring (starts n=3)2 planes$-2 \dots +2$
f3Complex (starts n=4)3 planes$-3 \dots +3$
Radial vs angular nodes
Radial nodes depend on both $n$ and $l$ (spherical surfaces); angular nodes depend only on $l$ (planes/cones). Quick check: 3d → 0 radial, 2 angular; 4s → 3 radial, 0 angular.

Electronic Configuration

Filling order (Aufbau / $(n+l)$ rule)

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p$$

Lower $(n+l)$ fills first; for equal $(n+l)$, lower $n$ fills first.

RuleStatement
AufbauFill orbitals in order of increasing energy
Pauli exclusionNo two electrons share all four quantum numbers (max 2 per orbital, opposite spins)
Hund’s ruleSingly fill degenerate orbitals with parallel spins before pairing
SubshellMax electrons
s2
p6
d10
f14

Must-know exceptions (extra stability of half / fully-filled subshells)

ElementZExpectedActual
Cr24[Ar] 4s² 3d⁴[Ar] 4s¹ 3d⁵
Cu29[Ar] 4s² 3d⁹[Ar] 4s¹ 3d¹⁰
Mo42[Kr] 5s² 4d⁴[Kr] 5s¹ 4d⁵
Ag47[Kr] 5s² 4d⁹[Kr] 5s¹ 4d¹⁰
Au79[Xe] 6s² 4f¹⁴ 5d⁹[Xe] 6s¹ 4f¹⁴ 5d¹⁰

Stable configurations: fully filled s², p⁶, d¹⁰, f¹⁴; half filled p³, d⁵, f⁷.

Ions and magnetism

  • Cations: remove electrons from the outermost shell first — for transition metals, 4s before 3d (in ions 3d lies below 4s). e.g. Fe ([Ar] 3d⁶ 4s²) → Fe²⁺ ([Ar] 3d⁶).
  • Anions: add electrons to the lowest available orbital. e.g. O ([He] 2s² 2p⁴) → O²⁻ ([Ne]).
  • Magnetism: unpaired electrons → paramagnetic; all paired → diamagnetic.
Aufbau diagonal

Diagonal arrow order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p.


Limitations of Bohr’s Model

graph TD
    A[Bohr's Model fails] --> B[Only single-electron species: H, He+, Li2+]
    A --> C[No fine structure of lines]
    A --> D[No Zeeman effect: magnetic field]
    A --> E[No Stark effect: electric field]
    A --> F[Violates Heisenberg uncertainty]
    A --> G[No relative line intensities]
    A --> H[No explanation of chemical bonding]

One-Glance Headline Formulas

$$E = \frac{hc}{\lambda} \quad\bullet\quad h\nu = \phi + KE_{max} \quad\bullet\quad r_n = \frac{0.529\,n^2}{Z}\,\text{Å} \quad\bullet\quad E_n = -\frac{13.6\,Z^2}{n^2}\,\text{eV}$$$$\frac{1}{\lambda} = R_H Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \quad\bullet\quad \lambda = \frac{h}{mv} \quad\bullet\quad \Delta x\,\Delta p \geq \frac{h}{4\pi} \quad\bullet\quad N = \frac{n(n-1)}{2}$$