Atomic Structure Formulas For JEE 2026
Atomic Structure is an important chapter in JEE Chemistry because it helps students understand how atoms are structured and how they behave. In this topic, students study the development of different atomic models, including Bohr’s model for hydrogen-like atoms. It also covers key concepts such as quantum numbers and electronic configuration, which explain how electrons are arranged around the nucleus and how atoms form different chemical properties.
The chapter further introduces the wave-mechanical model of the atom, which provides a deeper explanation of electron movement and energy levels. Many questions in JEE are based on spectral series, quantum numbers, and electron configurations, so understanding these formulas is very important. For quick and effective revision, students can also refer to a well-organized JEE Mains Chemistry Formula PDF to review important formulas and concepts during their preparation.
Early Atomic Models
Definition: Atom
The smallest particle of an element that can take part in a chemical reaction.
Rutherford's Nuclear Model
- The atom has a tiny, dense, positively charged nucleus at the centre.
- The nucleus contains nearly all the mass of the atom.
- Electrons revolve around the nucleus in circular orbits (like planets around the Sun).
- Size of nucleus $$\approx 10^{-15}$$ m; size of atom $$\approx 10^{-10}$$ m (nucleus is $$\sim 10^{5}$$ times smaller).
Important Note
Rutherford's model had a fatal flaw: according to classical electromagnetic theory, a revolving electron should continuously radiate energy and spiral into the nucleus. This would mean atoms are unstable — but they clearly are stable! Bohr resolved this problem.
Bohr's Atomic Model
Definition: Hydrogen-like Species
An atom or ion with only one electron orbiting the nucleus, e.g., H ($$Z=1$$), He$$^+$$ ($$Z=2$$), Li$$^{2+}$$ ($$Z=3$$).
Bohr's Postulates
- Electrons revolve in certain fixed circular orbits called stationary states. While in these orbits, they do not radiate energy.
- Only those orbits are allowed where the angular momentum is an integer multiple of $$\frac{h}{2\pi}$$: $$$mvr = n\frac{h}{2\pi}, \quad n = 1, 2, 3, \ldots$$$
- When an electron jumps from one orbit to another, it absorbs or emits a photon: $$$\Delta E = E_2 - E_1 = h\nu$$$
Bohr Model Formulas
Radius of $$n$$-th Orbit
$$$r_n = \frac{n^2}{Z} \times a_0$$$
where $$a_0 = 0.529 \; \text{Å} = 0.529 \times 10^{-10} \; \text{m}$$ is the Bohr radius.
Key dependence: $$r_n \propto \dfrac{n^2}{Z}$$
Velocity of Electron in $$n$$-th Orbit
$$$v_n = \frac{Z}{n} \times v_0$$$
where $$v_0 = 2.18 \times 10^6 \; \text{m/s}$$ is the velocity in the first orbit of hydrogen.
Key dependence: $$v_n \propto \dfrac{Z}{n}$$
Energy of Electron in $$n$$-th Orbit
$$$E_n = -\frac{Z^2}{n^2} \times 13.6 \; \text{eV}$$$
The negative sign means the electron is bound to the nucleus. $$n = 1$$ is the most tightly bound (ground state). As $$n \to \infty$$, $$E \to 0$$ (free electron).
Key dependence: $$E_n \propto -\dfrac{Z^2}{n^2}$$
Worked Example: Li$$^{2+}$$ Third Orbit
Find the radius and energy of an electron in the 3rd orbit of Li$$^{2+}$$ ($$Z = 3$$).
Radius: $$r_3 = \dfrac{3^2}{3} \times 0.529 \; \text{Å} = 3 \times 0.529 = 1.587 \; \text{Å}$$
Energy: $$E_3 = -\dfrac{3^2}{3^2} \times 13.6 \; \text{eV} = -13.6 \; \text{eV}$$
Worked Example: Photon Wavelength
Calculate the wavelength of the photon emitted when an electron in hydrogen falls from $$n = 3$$ to $$n = 2$$.
$$\Delta E = 13.6\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6 \times \dfrac{5}{36} = 1.89 \; \text{eV}$$
Convert to joules: $$1.89 \times 1.6 \times 10^{-19} = 3.024 \times 10^{-19} \; \text{J}$$
$$\lambda = \dfrac{hc}{\Delta E} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.024 \times 10^{-19}} = 6.58 \times 10^{-7} \; \text{m} = 658 \; \text{nm}$$
This is in the visible (red) region — part of the Balmer series.
Tip: For quick JEE calculations, remember: $$E_1(\text{H}) = -13.6$$ eV, $$a_0 = 0.529$$ Å, $$v_0 = 2.18 \times 10^6$$ m/s. For any hydrogen-like atom, just multiply/divide by $$Z$$ and $$n$$ as per the formulas.
Spectral Series of Hydrogen
Rydberg Formula for Spectral Lines
$$$\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$$
where $$R_H = 1.097 \times 10^7 \; \text{m}^{-1}$$ is the Rydberg constant, $$n_1 < n_2$$.
Hydrogen Spectral Series
| Series | Transition to ($$n_1$$) | From ($$n_2$$) | Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, … | Ultraviolet |
| Balmer | 2 | 3, 4, 5, … | Visible |
| Paschen | 3 | 4, 5, 6, … | Infrared |
| Brackett | 4 | 5, 6, 7, … | Infrared |
| Pfund | 5 | 6, 7, 8, … | Far Infrared |
Number of Spectral Lines
If an electron is in the $$n$$-th orbit and falls to the ground state, the total number of possible spectral lines is:
$$$\text{Number of lines} = \frac{n(n-1)}{2}$$$
Worked Example: Spectral Lines
An electron in hydrogen is excited to $$n = 4$$. How many spectral lines can be emitted?
Number of lines $$= \dfrac{4(4-1)}{2} = \dfrac{4 \times 3}{2} = 6$$
The possible transitions are: $$4 \to 3$$, $$4 \to 2$$, $$4 \to 1$$, $$3 \to 2$$, $$3 \to 1$$, $$2 \to 1$$.
Tip: Remember the series names in order: Lyman, Balmer, Paschen, Brackett, Pfund. Mnemonic: Look Before People Bump Past. Only the Balmer series is visible — this is a favourite JEE fact.
Dual Nature of Matter
de Broglie Wavelength
$$$\lambda = \frac{h}{mv} = \frac{h}{p}$$$
where $$h = 6.63 \times 10^{-34} \; \text{J·s}$$, $$m$$ = mass, $$v$$ = velocity, $$p = mv$$ = momentum.
For an electron accelerated through a potential difference $$V$$:
$$$\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}} \; \text{Å} \quad (\text{for electrons})$$$
Worked Example: de Broglie Wavelength
Find the de Broglie wavelength of an electron moving at $$10^6$$ m/s.
$$\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} = \dfrac{6.63 \times 10^{-34}}{9.1 \times 10^{-25}} = 7.28 \times 10^{-10} \; \text{m} = 7.28 \; \text{Å}$$
This is comparable to atomic dimensions, which is why electron diffraction can be observed.
Heisenberg's Uncertainty Principle
Heisenberg's Uncertainty Principle
$$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$$
where $$\Delta x$$ = uncertainty in position, $$\Delta p = m \cdot \Delta v$$ = uncertainty in momentum.
Equivalently:
$$$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}$$$
Worked Example: Uncertainty Principle
If the uncertainty in position of an electron is $$10^{-10}$$ m, find the minimum uncertainty in its velocity.
$$\Delta v \geq \dfrac{h}{4\pi \cdot m \cdot \Delta x} = \dfrac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-10}} = 5.8 \times 10^{5} \; \text{m/s}$$
Photoelectric Effect
Einstein's Photoelectric Equation
$$$h\nu = \phi + \frac{1}{2}mv_{\max}^2$$$
Energy of incident photon = work function + maximum kinetic energy of ejected electron.
Equivalently:
$$$KE_{\max} = h\nu - \phi = h(\nu - \nu_0)$$$
where $$\nu_0 = \dfrac{\phi}{h}$$ is the threshold frequency.
Worked Example: Photoelectric Effect
The work function of sodium is 2.3 eV. Light of wavelength 300 nm falls on sodium. Find the maximum KE of the photoelectrons.
Energy of photon: $$E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.63 \times 10^{-19} \; \text{J}$$
Converting to eV: $$\dfrac{6.63 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.14 \; \text{eV}$$
$$KE_{\max} = 4.14 - 2.3 = \textbf{1.84 eV}$$
Tip: Quick conversion: $$E \; (\text{in eV}) = \dfrac{12400}{\lambda \; (\text{in Å})} = \dfrac{1240}{\lambda \; (\text{in nm})}$$. This saves time in JEE.
Quantum Numbers
The Four Quantum Numbers
| Name | Symbol | Values | Physical Meaning |
|---|---|---|---|
| Principal | $$n$$ | $$1, 2, 3, \ldots$$ | Size and energy of the shell |
| Azimuthal | $$l$$ | $$0, 1, \ldots, (n-1)$$ | Shape of orbital: $$l=0$$ (s), $$1$$ (p), $$2$$ (d), $$3$$ (f) |
| Magnetic | $$m_l$$ | $$-l$$ to $$+l$$ | Orientation of orbital; total $$(2l+1)$$ values |
| Spin | $$m_s$$ | $$+\frac{1}{2}$$ or $$-\frac{1}{2}$$ | Spin direction ("up" or "down") |
Maximum Electrons in a Shell or Subshell
- Maximum electrons in a shell with quantum number $$n$$: $$2n^2$$
- Maximum electrons in a subshell with quantum number $$l$$: $$2(2l+1)$$
- s-subshell: 2 electrons, p-subshell: 6, d-subshell: 10, f-subshell: 14
Worked Example: Quantum Numbers
For $$n = 3$$, list all possible values of $$l$$, $$m_l$$, and the corresponding orbitals.
$$l = 0, 1, 2$$ (three subshells: 3s, 3p, 3d)
For $$l = 0$$: $$m_l = 0$$ → 1 orbital (3s)
For $$l = 1$$: $$m_l = -1, 0, +1$$ → 3 orbitals (3p)
For $$l = 2$$: $$m_l = -2, -1, 0, +1, +2$$ → 5 orbitals (3d)
Total orbitals in $$n = 3$$: $$1 + 3 + 5 = 9 = n^2$$. Total electrons: $$2 \times 9 = 18 = 2n^2$$.
Shapes of Orbitals and Nodes
Orbital Shapes and Characteristics
| Subshell | $$l$$ | Shape | No. of Nodal Planes |
|---|---|---|---|
| s | 0 | Spherical | 0 |
| p | 1 | Dumbbell (two lobes) | 1 |
| d | 2 | Double dumbbell / cloverleaf | 2 |
| f | 3 | Complex multi-lobed | 3 |
Nodes in Orbitals
- Radial (spherical) nodes = $$n - l - 1$$
- Angular (planar) nodes = $$l$$
- Total nodes = $$n - 1$$
Worked Example: Nodes
Find the number of radial and angular nodes for a 3p orbital.
For 3p: $$n = 3$$, $$l = 1$$.
Radial nodes $$= n - l - 1 = 3 - 1 - 1 = 1$$
Angular nodes $$= l = 1$$
Total nodes $$= n - 1 = 2$$ ✓
Electronic Configuration
Order of Filling Orbitals (Aufbau Order)
Orbitals fill in order of increasing $$(n + l)$$ value. If two orbitals have the same $$(n + l)$$, the one with lower $$n$$ fills first.
$$$1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p \to 5s \to 4d \to 5p \to 6s \to 4f \to 5d \to 6p \to 7s \to 5f \to 6d \to 7p$$$
Pauli's Exclusion Principle
No two electrons in an atom can have all four quantum numbers identical. This means each orbital can hold a maximum of 2 electrons, and they must have opposite spins.
Hund's Rule of Maximum Multiplicity
When filling orbitals of equal energy (degenerate orbitals), electrons first occupy each orbital singly with parallel spins before pairing up.
Worked Example: Electronic Configuration
Write the electronic configuration of Fe ($$Z = 26$$).
Following the Aufbau order:
$$1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 4s^2 \; 3d^6$$
Total electrons: $$2 + 2 + 6 + 2 + 6 + 2 + 6 = 26$$ ✓
Shorthand using noble gas core: $$[\text{Ar}] \; 4s^2 \; 3d^6$$
Important: Exceptions to Aufbau Principle
Exceptions occur at $$d^4$$ and $$d^9$$ configurations because half-filled ($$d^5$$) and fully-filled ($$d^{10}$$) d-subshells are extra stable:
- Chromium (Cr, $$Z = 24$$): Expected $$[\text{Ar}] \; 4s^2 \; 3d^4$$ → Actual: $$[\text{Ar}] \; 4s^1 \; 3d^5$$
- Copper (Cu, $$Z = 29$$): Expected $$[\text{Ar}] \; 4s^2 \; 3d^9$$ → Actual: $$[\text{Ar}] \; 4s^1 \; 3d^{10}$$
Tip: In JEE, you may be asked to find the number of unpaired electrons. Write the electronic configuration, then apply Hund's rule. For example, Fe$$^{2+}$$ ($$[\text{Ar}] \; 3d^6$$) has 4 unpaired electrons — the 5 d-orbitals get one electron each first (5 unpaired), then the 6th electron pairs with one (leaving 4 unpaired).
Important Constants for Atomic Structure
Key Constants
| Constant | Value |
|---|---|
| Planck's constant, $$h$$ | $$6.626 \times 10^{-34} \; \text{J·s}$$ |
| Mass of electron, $$m_e$$ | $$9.109 \times 10^{-31} \; \text{kg}$$ |
| Charge of electron, $$e$$ | $$1.602 \times 10^{-19} \; \text{C}$$ |
| Bohr radius, $$a_0$$ | $$0.529 \; \text{Å} = 52.9 \; \text{pm}$$ |
| Rydberg constant, $$R_H$$ | $$1.097 \times 10^7 \; \text{m}^{-1}$$ |
| Speed of light, $$c$$ | $$3 \times 10^8 \; \text{m/s}$$ |
| Ground state energy of H | $$-13.6 \; \text{eV}$$ |
Atomic Structure Formulas For JEE 2026: Conclusion
Atomic Structure is one of the most fundamental chapters in chemistry because it explains how atoms are organized and how electrons behave within them. Concepts such as Bohr’s model, quantum numbers, electronic configuration, and spectral series provide a deeper understanding of atomic behavior and energy levels.
By regularly revising the important formulas and practicing related numerical problems, students can build a strong conceptual foundation in this topic. A clear understanding of these principles helps in solving many physics and chemistry problems in competitive exams and strengthens overall problem-solving skills.
