Atomic Structure JEE Notes PDF, Formulas, Practice Questions

Atomic Structure carries some of the most formula-heavy and frequently asked questions in JEE Chemistry. It also forms the foundation for periodic properties, chemical bonding and coordination compounds. An atom is the smallest particle of an element that can take part in a chemical reaction, and our understanding of atomic structure developed through a series of models. Practising JEE questions on Atomic Structure helps students understand how concepts such as Bohr's model, quantum numbers, spectral series and electronic configuration are applied in numerical and conceptual problems. These Atomic Structure JEE notes cover early atomic models, Bohr's model, hydrogen spectral series, de Broglie wavelength, Heisenberg's uncertainty principle, photoelectric effect, quantum numbers, orbitals and electronic configuration for quick JEE revision.

Atomic Structure JEE Notes: Early Atomic Models

Dalton's Atomic Model

John Dalton proposed one of the earliest scientific atomic theories in 1803. According to Dalton, matter is made of extremely small particles called atoms.

• All matter is made of indivisible atoms.
• Atoms of the same element are identical in mass and properties.
• Atoms of different elements have different masses and properties.
• Atoms combine in simple whole-number ratios to form compounds.
• Atoms are neither created nor destroyed in a chemical reaction; they are only rearranged.

Limitation: Dalton's atomic theory could not explain the later discovery of subatomic particles such as electrons, protons and neutrons. It also could not explain isotopes, which are atoms of the same element with different masses.

Thomson's Atomic Model

J.J. Thomson discovered the electron through cathode ray experiments and proposed the plum pudding model. According to this model, the atom was a positively charged sphere with negatively charged electrons embedded inside it.

• Charge of electron: $$e = 1.6 \times 10^{-19}\,C$$
• Mass of electron: $$m_e = 9.1 \times 10^{-31}\,kg$$

Limitation: Thomson's model could explain the overall electrical neutrality of an atom but failed to explain the observations made in Rutherford's alpha-particle scattering experiment.

Rutherford's Atomic Model

Rutherford bombarded a thin gold foil with positively charged alpha particles and observed how they passed through or were deflected by the foil.

Rutherford's nuclear model:

• A tiny, dense and positively charged nucleus is present at the centre of the atom.
• Nearly all the mass of the atom is concentrated in the nucleus.
• Electrons revolve around the nucleus.
• The size of the nucleus is approximately $$10^{-15}\,m$$, while the size of an atom is approximately $$10^{-10}\,m$$.

Major limitation: According to classical electromagnetic theory, a revolving charged electron should continuously radiate energy, lose energy and eventually spiral into the nucleus. Rutherford's model therefore could not explain why atoms are stable.

Bohr's Model and Hydrogen Spectral Series

Bohr's Postulates

Niels Bohr proposed his atomic model in 1913 to explain atomic stability and the line spectrum of hydrogen. His model works for hydrogen-like species containing only one electron, such as H, He⁺ and Li²⁺.

• Electrons revolve around the nucleus only in certain permitted circular orbits called stationary states.
• An electron does not radiate energy while remaining in an allowed orbit.
• The angular momentum of an electron is quantised.

$$mvr = \frac{nh}{2\pi}$$

Here, $$n = 1, 2, 3, \ldots$$ is the principal quantum number.

When an electron moves from one energy level to another, energy is absorbed or emitted as a photon.

$$\Delta E = E_2-E_1 = h\nu$$

Bohr Model Formulas

For a hydrogen-like species of atomic number $$Z$$ and an electron present in the $$n^{th}$$ orbit:

The negative sign of energy indicates that the electron is bound to the nucleus. The $$n=1$$ level is the ground state and is the most tightly bound state. As $$n$$ approaches infinity, energy approaches zero and the electron becomes free.

Worked example: calculate the radius and energy of an electron in the third orbit of $$Li^{2+}$$, where $$Z=3$$.

$$r_3 = \frac{3^2}{3}\times0.529 = 3\times0.529 = 1.587\,\text{Å}$$

$$E_3 = -\frac{3^2}{3^2}\times13.6 = -13.6\,\text{eV}$$

JEE tip: Remember the three important hydrogen values: $$E_1=-13.6\,eV$$, $$a_0=0.529\,Å$$ and $$v_0=2.18\times10^6\,m/s$$. Values for other hydrogen-like species can be obtained by scaling them using $$Z$$ and $$n$$.

Hydrogen Spectral Series

When an excited electron in a hydrogen atom falls from a higher energy level to a lower energy level, it emits radiation of a specific wavelength. Different final energy levels produce different spectral series.

$$\frac{1}{\lambda}=R_HZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

Here, $$R_H = 1.097\times10^7\,m^{-1}$$ and $$n_2>n_1$$.

Number of possible spectral lines:

$$N = \frac{n(n-1)}{2}$$

Worked example: an electron excited to $$n=4$$ can produce:

$$N = \frac{4(4-1)}{2} = 6$$

The possible transitions are 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1.

JEE tip: Only the Balmer series lies in the visible region. The Lyman series lies in ultraviolet, while Paschen, Brackett and Pfund lie in infrared regions.

Dual Nature, Uncertainty Principle and Photoelectric Effect

de Broglie Wavelength

Louis de Broglie proposed that moving matter can show wave-like behaviour. Therefore, particles such as electrons possess a wavelength called the de Broglie wavelength.

$$\lambda = \frac{h}{mv} = \frac{h}{p}$$

Here, $$h = 6.63\times10^{-34}\,J\,s$$.

For an electron accelerated through a potential difference $$V$$:

$$\lambda = \frac{12.27}{\sqrt{V}}\,\text{Å}$$

Worked example: find the de Broglie wavelength of an electron moving at $$10^6\,m/s$$.

$$\lambda = \frac{6.63\times10^{-34}}{(9.1\times10^{-31})(10^6)}$$

$$\lambda = 7.28\times10^{-10}\,m = 7.28\,Å$$

Heisenberg's Uncertainty Principle

Heisenberg's uncertainty principle states that it is impossible to determine both the exact position and exact momentum of a microscopic particle at the same time.

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

Since $$p=mv$$:

$$\Delta x\cdot m\Delta v\geq\frac{h}{4\pi}$$

This principle is one of the reasons electrons cannot be described as travelling along exact fixed paths. Modern atomic theory instead describes electrons through regions of high probability called orbitals.

Photoelectric Effect

The photoelectric effect occurs when light of sufficiently high frequency strikes a metal surface and ejects electrons. Einstein explained the phenomenon by considering light as packets of energy called photons.

Energy of one photon:

$$E = h\nu$$

The minimum amount of energy required to eject an electron from a metal surface is called the work function, represented by $$\phi$$.

$$h\nu = \phi + \frac{1}{2}mv_{\max}^2$$

Therefore:

$$KE_{\max} = h\nu-\phi = h(\nu-\nu_0)$$

where:

$$\nu_0 = \frac{\phi}{h}$$

is the threshold frequency.

Worked example: sodium has a work function of 2.3 eV and is illuminated by light of wavelength 300 nm.

• Photon energy = $$\frac{1240}{300} = 4.13\,eV$$ approximately.
• $$KE_{\max} = 4.13-2.3 = 1.83\,eV$$ approximately.

JEE tip: For quick photon-energy calculations, use:

$$E(\text{eV}) = \frac{12400}{\lambda(\text{Å})} = \frac{1240}{\lambda(\text{nm})}$$

Quantum Numbers, Orbitals and Electronic Configuration

The Four Quantum Numbers

The quantum mechanical model uses four quantum numbers to describe the state of an electron inside an atom.

Worked example: consider $$n=3$$.

• Possible $$l$$ values are 0, 1 and 2, corresponding to 3s, 3p and 3d.
• For $$l=0$$, $$m_l=0$$, giving 1 orbital.
• For $$l=1$$, $$m_l=-1,0,+1$$, giving 3 orbitals.
• For $$l=2$$, $$m_l=-2,-1,0,+1,+2$$, giving 5 orbitals.
• Total orbitals = $$1+3+5=9=n^2$$.
• Maximum electrons = $$18=2n^2$$.

Maximum Electron Capacity

• Maximum electrons in a shell = $$2n^2$$.
• Maximum electrons in a subshell = $$2(2l+1)$$.
• s subshell can contain 2 electrons.
• p subshell can contain 6 electrons.
• d subshell can contain 10 electrons.
• f subshell can contain 14 electrons.

Shapes of Orbitals and Nodes

An orbital is a region around the nucleus where the probability of finding an electron is high. Unlike the fixed paths in Bohr's model, orbitals represent probability distributions.

Node formulas:

• Radial nodes = $$n-l-1$$
• Angular nodes = $$l$$
• Total nodes = $$n-1$$

Worked example: for a 3p orbital, $$n=3$$ and $$l=1$$.

• Radial nodes = $$3-1-1=1$$.
• Angular nodes = $$1$$.
• Total nodes = $$3-1=2$$.

Rules of Electronic Configuration

Aufbau principle: electrons occupy orbitals in order of increasing energy. Orbitals generally fill according to increasing $$(n+l)$$ values. If two orbitals have the same $$(n+l)$$ value, the orbital with lower $$n$$ fills first.

Common filling order:

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p

Pauli's exclusion principle: no two electrons in an atom can have identical values of all four quantum numbers. Therefore, one orbital can contain a maximum of two electrons, and they must have opposite spins.

Hund's rule of maximum multiplicity: electrons occupy degenerate orbitals singly with parallel spins before pairing begins.

Worked example: electronic configuration of Fe, $$Z=26$$:

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$$

Shorthand configuration:

$$[Ar]\,4s^2\,3d^6$$

Chromium and Copper Exceptions

Half-filled and completely filled d subshells have additional stability. Therefore, chromium and copper show exceptions to their expected Aufbau configurations.

JEE tip: To find the number of unpaired electrons, first write the electronic configuration and then distribute electrons according to Hund's rule.

Atomic Structure Formula Sheet at a Glance

Important Constants for Atomic Structure

JEE Important Points, Common Mistakes and Quick Revision

Points JEE Repeatedly Tests

• Bohr's formulas apply only to hydrogen-like species containing exactly one electron, such as H, He⁺ and Li²⁺.
• Energy varies as $$-\frac{Z^2}{n^2}$$, radius as $$\frac{n^2}{Z}$$ and velocity as $$\frac{Z}{n}$$.
• Only the Balmer series lies in the visible region.
• Lyman lies in ultraviolet, while Paschen, Brackett and Pfund lie in infrared regions.
• Aufbau filling follows the $$(n+l)$$ rule.
• Chromium and copper are important electronic configuration exceptions.
• The uncertainty principle replaces the idea of fixed electron paths with probability-based orbitals.
• The energy of a bound electron is negative and approaches zero as $$n$$ increases.

Common Mistakes to Avoid

1. Forgetting the Z or Z² factor. For hydrogen-like species, energy depends on $$Z^2$$, radius on $$1/Z$$ and velocity on $$Z$$.
2. Dropping the negative sign in Bohr energy. The negative sign indicates that the electron is bound to the nucleus.
3. Reversing n₁ and n₂ in the Rydberg equation. The lower energy level must be $$n_1$$ and the higher energy level must be $$n_2$$.
4. Confusing radial and angular nodes. Radial nodes are $$n-l-1$$, while angular nodes equal $$l$$.
5. Applying Bohr formulas to multi-electron atoms. These formulas are intended for hydrogen-like one-electron species.
6. Using the wrong Aufbau filling order. The 4s orbital fills before the 3d orbital in neutral atoms.
7. Forgetting unit conversions. Convert eV into joules and wavelength units correctly whenever the formula requires SI units.
8. Assuming higher light intensity increases photoelectron kinetic energy. Intensity mainly affects the number of emitted photoelectrons, while frequency determines their maximum kinetic energy.

Quick Revision Notes for Atomic Structure

• Atomic model sequence: Dalton → Thomson → Rutherford → Bohr → quantum mechanical model.
• Bohr formulas: $$r_n=\frac{n^2}{Z}(0.529\,Å)$$, $$v_n=\frac{Z}{n}(2.18\times10^6\,m/s)$$ and $$E_n=-\frac{Z^2}{n^2}(13.6\,eV)$$.
• Rydberg formula: $$\frac1\lambda=R_HZ^2\left(\frac1{n_1^2}-\frac1{n_2^2}\right)$$.
• Maximum number of possible spectral lines from level $$n$$ is $$\frac{n(n-1)}2$$.
• Spectral series order: Lyman, Balmer, Paschen, Brackett and Pfund. Only Balmer lies in the visible region.
• de Broglie wavelength: $$\lambda=\frac{h}{mv}$$.
• Uncertainty relation: $$\Delta x\Delta p\geq\frac{h}{4\pi}$$.
• Photoelectric equation: $$KE_{\max}=h\nu-\phi$$.
• Quantum numbers: $$n$$ represents shell, $$l$$ represents subshell and shape, $$m_l$$ represents orientation and $$m_s$$ represents spin.
• Maximum electrons in a shell = $$2n^2$$; maximum electrons in a subshell = $$2(2l+1)$$.
• Radial nodes = $$n-l-1$$, angular nodes = $$l$$ and total nodes = $$n-1$$.
• Electronic configuration follows Aufbau, Pauli and Hund's rules, with chromium and copper as important exceptions.

Problem-solving routine: First identify whether the question belongs to the Bohr model, hydrogen spectrum, matter waves, photoelectric effect, quantum numbers, orbitals or electronic configuration. Write the relevant relation and carefully check values such as Z, n and l along with the units before calculating. Use a JEE formula sheet during revision to quickly recall important Atomic Structure formulas, constants and relationships.