What are the four quantum numbers?

The four quantum numbers are: Principal quantum number (n) - determines shell and energy, Azimuthal quantum number (l) - determines subshell and shape of orbital, Magnetic quantum number (m) - determines orientation of orbital, and Spin quantum number (s) - determines spin direction of electron (+1/2 or -1/2).

What is the formula for energy of electron in Bohr's model?

The energy of electron in nth orbit is given by: E_n = -13.6 Z²/n² eV, where Z is atomic number and n is principal quantum number. For hydrogen atom (Z=1), ground state energy (n=1) is -13.6 eV.

What is Aufbau principle?

Aufbau principle states that electrons fill orbitals starting from lowest energy to highest energy. The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. The (n+l) rule helps determine the order.

What is de Broglie wavelength?

de Broglie wavelength is the wavelength associated with a moving particle. It is given by λ = h/mv = h/p, where h is Planck's constant, m is mass, v is velocity, and p is momentum. This concept establishes wave-particle duality of matter.

What are the limitations of Bohr's model?

Limitations of Bohr's model include: (1) Cannot explain spectra of multi-electron atoms, (2) Cannot explain Zeeman and Stark effects, (3) Cannot explain fine structure of spectral lines, (4) Violates Heisenberg uncertainty principle, (5) Cannot explain chemical bonding.

How many electrons can each subshell hold?

Each subshell can hold maximum 2(2l+1) electrons: s-subshell (l=0) holds 2 electrons, p-subshell (l=1) holds 6 electrons, d-subshell (l=2) holds 10 electrons, f-subshell (l=3) holds 14 electrons.

Atomic Structure for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Discovery of Subatomic Particles

The atom, once thought to be indivisible, was shown to contain smaller particles. Understanding the discovery experiments is crucial for JEE as conceptual questions are frequently asked.

1.1 Discovery of Electron (J.J. Thomson, 1897)

Cathode Ray Experiment

Key Observations:

Rays travel in straight lines from cathode to anode
Rays are deflected by electric and magnetic fields
Rays produce fluorescence and heating effect
Properties independent of electrode material and gas used

Charge to Mass Ratio (e/m):

\frac{e}{m} = 1.758820 \times 10^{11} \text{ C/kg}

Property	Electron	Proton	Neutron
Symbol	e⁻ or ⁰₋₁e	p⁺ or ¹₁H	n or ¹₀n
Discoverer	J.J. Thomson (1897)	Goldstein (1886), Rutherford (1919)	James Chadwick (1932)
Charge	-1.6 × 10⁻¹⁹ C	+1.6 × 10⁻¹⁹ C	0 (neutral)
Mass (kg)	9.109 × 10⁻³¹	1.673 × 10⁻²⁷	1.675 × 10⁻²⁷
Mass (u)	0.000549 u	1.00727 u	1.00867 u
Location	Outside nucleus (orbitals)	Inside nucleus	Inside nucleus

1.2 Millikan's Oil Drop Experiment (1909)

Purpose: Determination of Charge of Electron

e = 1.6022 \times 10^{-19} \text{ C}

Millikan found that all charges were multiples of a basic unit, proving charge is quantized.

From e/m and e values:

m_e = \frac{e}{e/m} = \frac{1.6 \times 10^{-19}}{1.76 \times 10^{11}} = 9.109 \times 10^{-31} \text{ kg}

1.3 Atomic Terminology

Atomic Number (Z)

Number of protons in nucleus

Z = Number of protons

For neutral atom: Z = Number of electrons

Mass Number (A)

Total nucleons (protons + neutrons)

A = Z + N

where N = number of neutrons

1.4 Isotopes, Isobars, Isotones

Term	Same	Different	Examples
Isotopes	Atomic number (Z)	Mass number (A)	¹H, ²H, ³H (Hydrogen isotopes)
Isobars	Mass number (A)	Atomic number (Z)	⁴⁰Ar, ⁴⁰K, ⁴⁰Ca
Isotones	Neutron number (N)	Z and A	¹³C, ¹⁴N, ¹⁵O (N=7)
Isoelectronic	Number of electrons	Z and A	Na⁺, Ne, F⁻ (10 e⁻)

💡 JEE Memory Trick

ISO-TOPES → Same TOP (atomic number), different bottom
ISO-BARS → Same BAR (mass number)
ISO-TONES → Same NEUTRONS

Atomic Models

Understanding the evolution of atomic models is crucial for JEE. Each model built upon previous discoveries and had specific limitations that led to newer models.

2.1 Thomson's Plum Pudding Model (1898)

Postulates

Atom is a sphere of positive charge
Electrons embedded in positive charge
Total positive charge = Total negative charge (neutral atom)
Also called "Watermelon model" or "Plum pudding model"

Limitations

Could not explain Rutherford's α-scattering experiment
Could not explain atomic spectra
Could not explain nuclear stability

2.2 Rutherford's Nuclear Model (1911)

α-Particle Scattering Experiment

Observations:

Most α-particles passed through undeflected
Some particles deflected at small angles
Very few (1 in 20,000) deflected by angles > 90°
Some particles (1 in 8000) bounced back at 180°

Conclusions & Postulates

Most of atom is empty space (most particles pass through)
Positive charge concentrated in a small dense region called nucleus
Nucleus contains most of the mass of atom
Electrons revolve around nucleus in circular orbits
Size of nucleus ≈ 10⁻¹⁵ m, Size of atom ≈ 10⁻¹⁰ m

Important Formula - Distance of Closest Approach:

r_0 = \frac{kZe \cdot 2e}{\frac{1}{2}m_\alpha v^2} = \frac{4kZe^2}{m_\alpha v^2}

where k = 9 × 10⁹ N·m²/C²

⚠️ Limitations of Rutherford's Model

Stability Problem: According to classical electrodynamics, accelerating electron should emit radiation and spiral into nucleus
Atomic Spectra: Could not explain line spectra of atoms (only discrete wavelengths)
Energy Problem: Could not explain why electrons don't lose energy continuously

Bohr's Model of Atom (1913)

Bohr's model is the MOST IMPORTANT topic in Atomic Structure for JEE. It explains hydrogen spectrum and provides formulas that are tested every year. Master all formulas!

3.1 Postulates of Bohr's Model

Postulate 1: Stationary Orbits

Electrons revolve around nucleus in certain fixed circular orbits called stationary orbits or energy levels without radiating energy.

Postulate 2: Quantization of Angular Momentum

Angular momentum of electron is quantized:

mvr = n\frac{h}{2\pi} = n\hbar

where n = 1, 2, 3, ... (principal quantum number), ℏ = h/2π

Postulate 3: Energy Transition

Electron absorbs or emits energy when it jumps between orbits:

\Delta E = E_2 - E_1 = h\nu = \frac{hc}{\lambda}

3.2 Bohr's Model Formulas (Must Memorize!)

🎯 Critical Formulas for JEE

1. Radius of nth Orbit

r_n = 0.529 \times \frac{n^2}{Z} \text{ Å}

For H atom (Z=1): r₁ = 0.529 Å (Bohr radius)

2. Velocity of Electron

v_n = 2.18 \times 10^6 \times \frac{Z}{n} \text{ m/s}

For H atom: v₁ = 2.18 × 10⁶ m/s

3. Energy of Electron

E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}

Also: Eₙ = -2.18 × 10⁻¹⁸ × Z²/n² J

4. Time Period

T_n = 1.52 \times 10^{-16} \times \frac{n^3}{Z^2} \text{ s}

T ∝ n³/Z²

5. Frequency of Revolution

\nu_n = 6.58 \times 10^{15} \times \frac{Z^2}{n^3} \text{ Hz}

ν ∝ Z²/n³

6. Current in Orbit

I = e\nu = \frac{ev}{2\pi r}

I ∝ Z²/n³

💡 Quick Memory: Proportionality Relations

r ∝ n²/Z

v ∝ Z/n

E ∝ Z²/n²

T ∝ n³/Z²

ν ∝ Z²/n³

KE ∝ Z²/n²

3.3 Energy Relations

Kinetic, Potential & Total Energy

Kinetic Energy

KE = +13.6 \times \frac{Z^2}{n^2} \text{ eV}

KE is always positive

Potential Energy

PE = -27.2 \times \frac{Z^2}{n^2} \text{ eV}

PE is always negative

Total Energy

TE = -13.6 \times \frac{Z^2}{n^2} \text{ eV}

TE = KE + PE

Important Relationships:

KE = -TE = -½PE

PE = 2TE = -2KE

📝 Solved Example 1 (JEE Main Pattern)

Question: Calculate the radius, velocity, and energy of electron in 3rd orbit of He⁺ ion.

Solution:

For He⁺: Z = 2, n = 3

1. Radius:

r_3 = 0.529 \times \frac{n^2}{Z} = 0.529 \times \frac{9}{2} = 2.38 \text{ Å}

2. Velocity:

v_3 = 2.18 \times 10^6 \times \frac{Z}{n} = 2.18 \times 10^6 \times \frac{2}{3}

v_3 = 1.45 \times 10^6 \text{ m/s}

3. Energy:

E_3 = -13.6 \times \frac{Z^2}{n^2} = -13.6 \times \frac{4}{9} = -6.04 \text{ eV}

3.4 Hydrogen Spectrum

Rydberg Formula

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

where R = Rydberg constant = 1.097 × 10⁷ m⁻¹ = 109677 cm⁻¹

Series	n₁ (Lower)	n₂ (Upper)	Region	First Line
Lyman	1	2, 3, 4, ...	UV	Lα (1→2): 121.6 nm
Balmer	2	3, 4, 5, ...	Visible	Hα (2→3): 656.3 nm (Red)
Paschen	3	4, 5, 6, ...	IR	Pα (3→4): 1875 nm
Brackett	4	5, 6, 7, ...	IR	Bα (4→5)
Pfund	5	6, 7, 8, ...	Far IR	Pfα (5→6)

💡 Memory Trick for Spectral Series

"Lovely Balmer Plays Brass Perfectly"
L-yman (n₁=1), B-almer (n₁=2), P-aschen (n₁=3), Br-ackett (n₁=4), Pf-und (n₁=5)

Region: "UVVIr" → UV (Lyman), Visible (Balmer), IR (rest)

Number of Spectral Lines

When electron falls from nth orbit to ground state:

\text{Total lines} = \frac{n(n-1)}{2}

Lines in a particular series: (n - n₁) where n₁ is lower level of that series

📝 Solved Example 2 (JEE Main 2023 Type)

Question: Calculate the wavelength of first line of Balmer series (Hα line) for hydrogen.

Solution:

For Balmer series: n₁ = 2

For first line (Hα): n₂ = 3

Using Rydberg formula:

\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right)

\frac{1}{\lambda} = R\left(\frac{9-4}{36}\right) = \frac{5R}{36}

\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7}

\lambda = 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm (Red light)}

3.5 Limitations of Bohr's Model

⚠️ Limitations (Important for JEE)

Applies only to single electron systems (H, He⁺, Li²⁺, etc.)
Cannot explain fine structure of spectral lines (doublets)
Cannot explain Zeeman effect (splitting in magnetic field)
Cannot explain Stark effect (splitting in electric field)
Violates Heisenberg's uncertainty principle
Cannot explain chemical bonding
Does not explain relative intensity of spectral lines

Quantum Numbers

Quantum numbers are like the "address" of an electron in an atom. Four quantum numbers uniquely identify each electron. This is one of the highest-scoring topics in JEE!

4.1 The Four Quantum Numbers

1. Principal Quantum Number (n)

Determines:

Main energy level (shell)
Size of orbital
Maximum distance from nucleus

Values & Names:

n = 1, 2, 3, 4, ... (positive integers)

n = 1 (K shell), 2 (L shell), 3 (M shell), 4 (N shell)...

Max electrons in shell = 2n²

2. Azimuthal/Angular Momentum Quantum Number (l)

Determines:

Shape of orbital
Subshell
Orbital angular momentum

Values:

l = 0, 1, 2, ... (n-1)

l = 0 (s), 1 (p), 2 (d), 3 (f)

Angular momentum = √(l(l+1)) × ℏ

Max electrons in subshell = 2(2l+1)

3. Magnetic Quantum Number (m or mₗ)

Determines:

Orientation of orbital in space
Number of orbitals in subshell
Behavior in magnetic field

Values:

m = -l, (-l+1), ... 0, ... (+l-1), +l

Total values = (2l+1)

Number of orbitals in subshell = (2l+1)

4. Spin Quantum Number (s or mₛ)

Determines:

Direction of spin of electron
Magnetic behavior of electron

Values:

s = +½ (↑, clockwise)

s = -½ (↓, anticlockwise)

Spin angular momentum = √(s(s+1)) × ℏ = √3/2 × ℏ

4.2 Summary Table

Subshell	l value	m values	No. of orbitals	Max electrons
s	0	0	1	2
p	1	-1, 0, +1	3	6
d	2	-2, -1, 0, +1, +2	5	10
f	3	-3, -2, -1, 0, +1, +2, +3	7	14

📝 Solved Example 3 (JEE Main Pattern)

Question: Find all possible quantum numbers for an electron in 3p orbital.

Solution:

For 3p orbital:

n = 3 (3rd shell)
l = 1 (p subshell)
m = -1, 0, +1 (three possible values)
s = +½ or -½ (two possible values)

Possible combinations: 6

(3, 1, -1, +½), (3, 1, -1, -½), (3, 1, 0, +½), (3, 1, 0, -½), (3, 1, +1, +½), (3, 1, +1, -½)

💡 JEE Quick Formulas

Number of subshells in nth shell = n
Number of orbitals in nth shell = n²
Maximum electrons in nth shell = 2n²
Maximum electrons in a subshell = 2(2l+1)
Number of orbitals in a subshell = (2l+1)

Shapes of Orbitals

An orbital is a three-dimensional region of space where the probability of finding an electron is maximum. Understanding orbital shapes is crucial for chemical bonding concepts.

5.1 Shapes of Different Orbitals

s Orbital (l = 0)

⬤

Spherically Symmetric

Shape: Spherical
One orientation only
No angular node
Radial nodes = (n - 1)

p Orbital (l = 1)

⬤━━⬤

Dumbbell Shape

Shape: Dumbbell (two lobes)
3 orientations: pₓ, p_y, p_z
Angular node = 1 (nodal plane)
Radial nodes = (n - 2)

d Orbital (l = 2)

🍀 (cloverleaf)

Double Dumbbell Shape

Shape: Double dumbbell (4 lobes) or dumbbell + ring
5 orientations: d_xy, d_yz, d_xz, d_x²-y², d_z²
Angular nodes = 2
Radial nodes = (n - 3)

f Orbital (l = 3)

Complex

8 lobes

Shape: Complex (8 lobes)
7 orientations
Angular nodes = 3
Radial nodes = (n - 4)

5.2 Nodes in Orbitals

Types of Nodes

Radial (Spherical) Nodes

\[= n - l - 1\]

Angular (Planar) Nodes

\[= l\]

Total Nodes

\[= n - 1\]

📝 Solved Example 4

Question: Find the number of radial and angular nodes in 4d orbital.

Solution:

For 4d: n = 4, l = 2

\text{Radial nodes} = n - l - 1 = 4 - 2 - 1 = 1

\text{Angular nodes} = l = 2

\text{Total nodes} = n - 1 = 3

Electronic Configuration

Electronic configuration describes the distribution of electrons in various orbitals. Three rules govern this distribution, and exceptions are frequently tested in JEE.

6.1 Rules for Electronic Configuration

1. Aufbau Principle

"Electrons fill orbitals in order of increasing energy"

Order of Filling (using n+l rule):

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

(n+l) Rule:

Lower (n+l) value → fills first

If (n+l) same → lower n fills first

2. Pauli's Exclusion Principle

"No two electrons in an atom can have all four quantum numbers identical"

Consequence:

Maximum 2 electrons per orbital
They must have opposite spins (↑↓)
Maximum electrons in shell = 2n²

3. Hund's Rule of Maximum Multiplicity

"Electrons prefer to occupy empty orbitals with parallel spins before pairing"

Example: Nitrogen (7 electrons)

1s² 2s² 2p³ →

↑ ↑ ↑

✓ Correct

1s² 2s² 2p³ →

↑↓ ↑

✗ Wrong

6.2 Important Electronic Configurations

Element	Z	Configuration	Notes
Hydrogen	1	1s¹	-
Carbon	6	1s² 2s² 2p²	-
Nitrogen	7	1s² 2s² 2p³	Half-filled (stable)
Oxygen	8	1s² 2s² 2p⁴	-
Chromium	24	[Ar] 3d⁵ 4s¹	Exception!
Copper	29	[Ar] 3d¹⁰ 4s¹	Exception!
Iron	26	[Ar] 3d⁶ 4s²	-
Zinc	30	[Ar] 3d¹⁰ 4s²	Completely filled

6.3 Exceptions to Aufbau Principle

⚠️ Important Exceptions (JEE Favorite!)

Reason: Half-filled and fully-filled subshells are extra stable due to:

Symmetrical electron distribution
Maximum exchange energy

Chromium (Z = 24)

Expected: [Ar] 3d⁴ 4s²

Actual: [Ar] 3d⁵ 4s¹ (half-filled d)

Copper (Z = 29)

Expected: [Ar] 3d⁹ 4s²

Actual: [Ar] 3d¹⁰ 4s¹ (fully-filled d)

Other exceptions: Mo, Ag, Au, Pd, etc.

6.4 Configuration of Ions

Rule for Ion Formation

When forming cations, electrons are removed from the outermost shell first (highest n value), NOT from the orbital that was filled last.

Example: Fe²⁺ and Fe³⁺

Fe (Z=26): [Ar] 3d⁶ 4s²

Fe²⁺: [Ar] 3d⁶ (remove 2 electrons from 4s, NOT 3d)

Fe³⁺: [Ar] 3d⁵ (remove 1 more from 3d)

📝 Solved Example 5 (JEE Main Pattern)

Question: Write electronic configuration of Cu²⁺ and find number of unpaired electrons.

Solution:

Cu (Z=29): [Ar] 3d¹⁰ 4s¹ (exception)

Cu²⁺: Remove 2 electrons from outermost

First remove from 4s¹ → 1 electron removed

Then remove from 3d¹⁰ → 1 electron removed

Cu²⁺: [Ar] 3d⁹

3d⁹ has 1 unpaired electron

Photoelectric Effect

The photoelectric effect demonstrated the particle nature of light (photons). Einstein's explanation earned him the Nobel Prize. This is a high-weightage topic with numerical problems frequently asked.

7.1 Phenomenon

What is Photoelectric Effect?

When light of sufficient frequency falls on a metal surface, electrons are emitted. These emitted electrons are called photoelectrons.

Key Observations:

Below threshold frequency (ν₀), no emission regardless of intensity
Above ν₀, electrons emitted instantaneously
Kinetic energy of electrons depends on frequency, not intensity
Number of electrons emitted depends on intensity

7.2 Einstein's Photoelectric Equation

The Fundamental Equation

h\nu = \phi + KE_{max}

h\nu = h\nu_0 + \frac{1}{2}mv_{max}^2

hν

Energy of incident photon

φ = hν₀

Work function (threshold energy)

½mv²ₘₐₓ

Maximum kinetic energy

Related Formulas

Energy of Photon

E = h\nu = \frac{hc}{\lambda}

Threshold Wavelength

\lambda_0 = \frac{hc}{\phi}

Stopping Potential

eV_0 = \frac{1}{2}mv_{max}^2

Energy in eV

E(eV) = \frac{12400}{\lambda(Å)}

📝 Solved Example 6 (JEE Main Pattern)

Question: Light of wavelength 4000 Å falls on a metal surface. If threshold wavelength is 5000 Å, find the maximum kinetic energy of emitted electrons in eV.

Solution:

Step 1: Calculate energies

E_{incident} = \frac{12400}{4000} = 3.1 \text{ eV}

\phi = \frac{12400}{5000} = 2.48 \text{ eV}

Step 2: Apply Einstein's equation

KE_{max} = E - \phi = 3.1 - 2.48

KE_{max} = 0.62 \text{ eV}

💡 Quick Formula: E(eV) = 12400/λ(Å)

This is a super useful shortcut for JEE! Works when wavelength is in Ångströms and you want energy in eV.

Also: E(eV) = 1240/λ(nm)

Wave-Particle Duality

De Broglie proposed that matter, like light, has both wave and particle properties. Heisenberg's uncertainty principle sets fundamental limits on our knowledge. These concepts form the foundation of quantum mechanics.

8.1 de Broglie Wavelength

de Broglie Hypothesis

"Every moving particle has an associated wave"

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

where λ = de Broglie wavelength, h = Planck's constant, m = mass, v = velocity, p = momentum

Important de Broglie Formulas

For Electron accelerated by V volts

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

In terms of Kinetic Energy

\lambda = \frac{h}{\sqrt{2mKE}}

In terms of Temperature (for gases)

\lambda = \frac{h}{\sqrt{3mkT}}

For electron in nth orbit

2\pi r_n = n\lambda

Circumference = n × de Broglie wavelengths

📝 Solved Example 7 (JEE Advanced Pattern)

Question: Calculate de Broglie wavelength of an electron accelerated through 100 V.

Solution:

Using the formula for electron accelerated through V volts:

\lambda = \frac{12.27}{\sqrt{V}} = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10}

\lambda = 1.227 \text{ Å}

8.2 Heisenberg's Uncertainty Principle

The Uncertainty Principle

"It is impossible to simultaneously determine both position and momentum of a particle with absolute precision"

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

or equivalently:

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

Δx

Uncertainty in position

Δp = mΔv

Uncertainty in momentum

h/4π

≈ 0.528 × 10⁻³⁴ J·s

Other Forms of Uncertainty Principle

Energy-Time Uncertainty

\Delta E \cdot \Delta t \geq \frac{h}{4\pi}

Angular versions

\Delta \theta \cdot \Delta L \geq \frac{h}{4\pi}

📝 Solved Example 8

Question: The uncertainty in position of an electron is 0.1 Å. Calculate the uncertainty in its velocity.
(mass of electron = 9.1 × 10⁻³¹ kg, h = 6.6 × 10⁻³⁴ J·s)

Solution:

Given: Δx = 0.1 Å = 0.1 × 10⁻¹⁰ m = 10⁻¹¹ m

Using Heisenberg's principle:

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

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

\Delta v \geq \frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-11}}

\Delta v \geq 5.77 \times 10^6 \text{ m/s}

💡 Significance of Uncertainty Principle

Significant for microscopic particles (electrons, protons)
Negligible for macroscopic objects (balls, cars)
Proves that definite orbits (Bohr model) cannot exist
Leads to concept of probability (orbitals instead of orbits)

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Bohr's Model Calculations: 35%
✓ Electronic Configuration: 25%
✓ Quantum Numbers: 15%
✓ Photoelectric Effect: 15%
✓ de Broglie & Uncertainty: 10%

JEE Advanced (Last 5 Years)

✓ Bohr Model + Spectral Lines: 30%
✓ Quantum Numbers & Orbitals: 25%
✓ Configuration Exceptions: 20%
✓ Wave-Particle Duality: 15%
✓ Multi-concept Problems: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-18 marks (3-5 questions)

Difficulty Level: Medium to High

Time Required: 3-4 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Calculate the radius and energy of electron in 2nd orbit of He⁺ ion.
Find the wavelength of first line of Lyman series for hydrogen atom.
Write electronic configuration of: (a) Fe (Z=26) (b) Cu (Z=29) (c) Cr (Z=24)
How many spectral lines are possible when electron falls from n=5 to ground state?
Find all four quantum numbers for 3rd electron in nitrogen atom.
Calculate de Broglie wavelength of electron moving with velocity 10⁶ m/s.
Light of wavelength 3000 Å falls on metal with work function 2.5 eV. Find KE of photoelectrons.
Find number of radial and angular nodes in 4p orbital.
Write electronic configuration of Fe²⁺ and Fe³⁺ ions.
Which has more energy: electron in n=2 of H or n=1 of He⁺?

Level 2: Intermediate (JEE Main/Advanced)

Calculate the ionization energy of Li²⁺ ion from ground state in eV and kJ/mol.
An electron in H-atom jumps from n=5 to n=2. Find wavelength of radiation emitted.
The ratio of kinetic energies of electron in 1st orbit of H and He⁺ is?
How many electrons in an atom can have n=4 and l=2?
The uncertainty in velocity of electron is 2.2 × 10⁶ m/s. Find minimum uncertainty in position.
Calculate stopping potential when light of 2000 Å falls on metal with threshold wavelength 2500 Å.
Find the orbit number in H-atom where velocity of electron is c/137 (c = speed of light).
The wavelength of radiation emitted when electron in He⁺ falls from n=4 to n=2 equals wavelength of radiation when electron in H jumps from n=? to n=2.
Calculate total number of orbitals having n+l ≤ 4.
An electron is accelerated through 100 V. Its de Broglie wavelength will be?

Level 3: Advanced (JEE Advanced/Olympiad)

In Bohr's model, the ratio of time period of electron in 2nd orbit of He⁺ to that in 3rd orbit of H is?
The wavelength of first line of Balmer series of H is 6563 Å. Calculate wavelength of first line of Lyman series.
If ionization energy of H-atom is 13.6 eV, calculate the energy required to excite electron from n=1 to n=4.
The angular momentum of electron in nth orbit is 4h/π. Find the energy of electron in this orbit for H-atom.
Calculate the wave number of spectral line with maximum wavelength in Paschen series of Li²⁺.
An electron and proton have same de Broglie wavelength. Which has greater velocity and by what factor?
The photon energy in units of eV for electromagnetic waves of wavelength 2 pm is?
In photoelectric effect, if kinetic energy of photoelectrons doubles when wavelength changes from 400 nm to 300 nm, find work function of metal.
How many electrons in Cu (Z=29) have (n+l) value equal to 3?
An electron in excited H-atom has angular momentum 3h/2π. It makes transition to first excited state. Find the wavelength of photon emitted.

Get Solutions PDF with Step-by-Step Working

📋 Quick Formula Sheet - Atomic Structure

Bohr Model Formulas

r_n = 0.529 × n²/Z Å

v_n = 2.18 × 10⁶ × Z/n m/s

E_n = -13.6 × Z²/n² eV

T_n = 1.52 × 10⁻¹⁶ × n³/Z² s

KE = +13.6 Z²/n² eV

PE = -27.2 Z²/n² eV

TE = KE + PE = -13.6 Z²/n² eV

Spectral Lines

1/λ = RZ²(1/n₁² - 1/n₂²)

R = 1.097 × 10⁷ m⁻¹

Total lines = n(n-1)/2

E = hν = hc/λ

E(eV) = 12400/λ(Å)

ΔE = E₂ - E₁ = hν

Quantum Numbers

n = 1, 2, 3, ... (shells)

l = 0 to (n-1) (subshells)

m = -l to +l (orbitals)

s = +½ or -½ (spin)

Max e⁻ in shell = 2n²

Max e⁻ in subshell = 2(2l+1)

Orbitals in shell = n²

Nodes in Orbitals

Radial nodes = n - l - 1

Angular nodes = l

Total nodes = n - 1

Angular momentum = √(l(l+1))ℏ

Spin momentum = √(s(s+1))ℏ

Photoelectric Effect

hν = φ + ½mv²_max

φ = hν₀ (work function)

KE_max = eV₀ (stopping potential)

λ₀ = hc/φ (threshold wavelength)

Wave-Particle Duality

λ = h/mv = h/p (de Broglie)

λ = 12.27/√V Å (for e⁻)

λ = h/√(2mKE)

Δx·Δp ≥ h/4π (Heisenberg)

Δx·mΔv ≥ h/4π

ΔE·Δt ≥ h/4π

📥 Download Printable Formula Sheet PDF

Year	JEE Main	JEE Advanced	Topic Focus
2024	3 Questions (12 marks)	4 Questions (15 marks)	Bohr model, Electronic configuration, de Broglie
2023	2 Questions (8 marks)	3 Questions (12 marks)	Spectral lines, Quantum numbers, Photoelectric
2022	3 Questions (12 marks)	5 Questions (18 marks)	H-like species comparison, Wave-particle duality
2021	2 Questions (8 marks)	3 Questions (11 marks)	Nodes in orbitals, Energy calculations
2020	2 Questions (8 marks)	4 Questions (14 marks)	Configuration exceptions, Heisenberg principle

📑 Quick Navigation - Atomic Structure

Atomic Structure JEE Main & Advanced 2025-26

Discovery of Subatomic Particles

1.1 Discovery of Electron (J.J. Thomson, 1897)

Cathode Ray Experiment

Key Observations:

Charge to Mass Ratio (e/m):

1.2 Millikan's Oil Drop Experiment (1909)

Purpose: Determination of Charge of Electron

1.3 Atomic Terminology

Atomic Number (Z)

Mass Number (A)

1.4 Isotopes, Isobars, Isotones

💡 JEE Memory Trick

Atomic Models

2.1 Thomson's Plum Pudding Model (1898)

Postulates

Limitations

2.2 Rutherford's Nuclear Model (1911)

α-Particle Scattering Experiment

Observations:

Conclusions & Postulates

Important Formula - Distance of Closest Approach:

⚠️ Limitations of Rutherford's Model

Bohr's Model of Atom (1913)

3.1 Postulates of Bohr's Model

Postulate 1: Stationary Orbits

Postulate 2: Quantization of Angular Momentum

Postulate 3: Energy Transition

3.2 Bohr's Model Formulas (Must Memorize!)

🎯 Critical Formulas for JEE

1. Radius of nth Orbit

2. Velocity of Electron

3. Energy of Electron

4. Time Period

5. Frequency of Revolution

6. Current in Orbit

💡 Quick Memory: Proportionality Relations

3.3 Energy Relations

Kinetic, Potential & Total Energy

Kinetic Energy

Potential Energy

Total Energy

Important Relationships:

📝 Solved Example 1 (JEE Main Pattern)

3.4 Hydrogen Spectrum

Rydberg Formula

💡 Memory Trick for Spectral Series

Number of Spectral Lines

📝 Solved Example 2 (JEE Main 2023 Type)

3.5 Limitations of Bohr's Model

⚠️ Limitations (Important for JEE)

Quantum Numbers

4.1 The Four Quantum Numbers

1. Principal Quantum Number (n)

Determines:

Values & Names:

2. Azimuthal/Angular Momentum Quantum Number (l)

Determines:

Values:

3. Magnetic Quantum Number (m or mₗ)

Determines:

Values:

4. Spin Quantum Number (s or mₛ)

Determines:

Values:

4.2 Summary Table

📝 Solved Example 3 (JEE Main Pattern)

💡 JEE Quick Formulas

Shapes of Orbitals

5.1 Shapes of Different Orbitals

s Orbital (l = 0)

p Orbital (l = 1)

d Orbital (l = 2)

f Orbital (l = 3)

5.2 Nodes in Orbitals

Types of Nodes

Radial (Spherical) Nodes

Angular (Planar) Nodes

Total Nodes