NUCLEAR PHYSICS

Atomic Physics Class 12 Notes for JEE Main | Complete Formula Sheet, Bohr Model & Solved Questions

PHYSICS NOTES • CLASS 12 • JEE MAIN

Atomic Physics: Complete Notes, Important Formulas & Solved Numericals

Master Bohr's Model, Hydrogen Spectrum and Energy-Level Transitions.

NCERT Based JEE Main Preparation Quick Revision

Welcome to our Class 12 Physics study guide! In this post, we will cover Atomic Physics for JEE Main in a simple and exam-oriented manner. Learn the important concepts, memorize the essential formulas, understand solved examples, and test your preparation with practice MCQs.

📚 Table of Contents

1. Introduction to Atomic Physics
2. Atomic Models
3. Bohr's Atomic Model
4. Important Formula Sheet
5. Hydrogen Spectrum
6. Solved Numerical Problems
7. JEE Main Preparation Tips
8. Practice MCQs with Answers
9. Quick Revision Notes

1. Introduction to Atomic Physics

Atomic Physics studies the structure of atoms, the arrangement of electrons, and the emission or absorption of electromagnetic radiation when electrons change their energy states.

Important topics for Class 12 and JEE Main include:

• Rutherford's alpha-particle scattering experiment
• Bohr's postulates and atomic energy levels
• Radius, velocity and energy of an electron in a hydrogen-like atom
• Hydrogen emission spectrum and spectral series
• Excitation energy, ionization energy and photon energy
• Numerical problems involving energy transitions

Exam Tip: Pay special attention to Bohr's formulas and energy-level transitions. Practice questions involving the initial and final energy levels carefully.

2. Atomic Models

A. Thomson's Atomic Model

Thomson proposed that an atom consists of a positively charged sphere with negatively charged electrons embedded in it. This model could not explain Rutherford's alpha-particle scattering observations.

B. Rutherford's Nuclear Model

Rutherford's gold foil experiment led to these conclusions:

• Most of the atom is empty space.
• Nearly all the positive charge and most of the mass are concentrated in a tiny nucleus.
• Electrons occupy the region around the nucleus.

Limitation: According to classical electrodynamics, an accelerating electron should radiate energy and spiral into the nucleus. Rutherford's model could not explain atomic stability or the discrete line spectrum of hydrogen.

C. Bohr's Atomic Model

Bohr introduced quantized electron orbits to explain the stability of hydrogen and its line spectrum.

3. Bohr's Atomic Model

Bohr's model is based on the following important postulates:

1. Electrons can occupy certain permitted stationary orbits without continuously radiating energy.
2. The angular momentum of an electron is quantized:
   mvr = nℏ = nh/(2π)
   where n = 1, 2, 3, ... is the principal quantum number.
3. Radiation is emitted or absorbed when an electron moves between two permitted energy levels.
   |ΔE| = hν = hc/λ

Radius of the nth Orbit

r_n = 0.529 × n²/Z Å

Here, Z is the atomic number of a hydrogen-like species containing only one electron. For hydrogen, Z = 1.

For hydrogen:

• First orbit: r₁ = 0.529 Å
• Second orbit: r₂ = 2.116 Å
• Third orbit: r₃ = 4.761 Å

Velocity of the Electron

v_n = 2.18 × 10⁶ (Z/n) m/s

The electron's speed decreases as n increases for a fixed hydrogen-like species.

Energy of the Electron

E_n = −13.6 Z²/n² eV

The negative sign indicates that the electron is in a bound state. The zero of energy is taken to be a free electron at infinity.

Orbit	n	Energy in Hydrogen
Ground state	1	−13.6 eV
First excited state	2	−3.4 eV
Second excited state	3	−1.51 eV approximately
Third excited state	4	−0.85 eV
Ionization limit	∞	0 eV

Remember: The ground state is the lowest-energy state. Energy becomes less negative as n increases and approaches zero at the ionization limit.

4. Important Atomic Physics Formulas for JEE Main

1. Energy of the nth orbit

E_n = −13.6 Z²/n² eV

2. Radius of the nth orbit

r_n = 0.529 n²/Z Å

3. Electron velocity

v_n = 2.18 × 10⁶ Z/n m/s

4. Kinetic energy

K_n = +13.6 Z²/n² eV

5. Potential energy

U_n = −27.2 Z²/n² eV

6. Total energy relationship

E = K + U = −K = U/2

7. Photon energy

E_photon = hν = hc/λ

8. Wavelength of a spectral line

1/λ = RZ² (1/n_f² − 1/n_i²)

For emission, n_i > n_f. The Rydberg constant R is approximately 1.097 × 10⁷ m⁻¹.

9. Ionization energy from the nth level

E_ionization = 13.6 Z²/n² eV

10. Excitation energy from the ground state

ΔE = 13.6 Z² (1 − 1/n²) eV

This formula applies when an electron moves from n = 1 to level n.

Common mistake: Use Z² in energy and Z in the denominator of the orbit-radius formula. Do not confuse the emitted photon's energy with the final electron's total energy.

5. Hydrogen Spectrum and Spectral Series

When an electron falls from a higher energy level to a lower level, the atom emits a photon. Each allowed transition corresponds to a particular wavelength.

Spectral Series	Final Level	Region
Lyman	nf = 1	Ultraviolet
Balmer	nf = 2	Visible and near-ultraviolet
Paschen	nf = 3	Infrared
Brackett	nf = 4	Infrared
Pfund	nf = 5	Infrared

Emission vs Absorption

Emission: Electron moves from a higher energy level to a lower one. A photon is released.

Absorption: Electron absorbs a suitable amount of energy and moves to a higher energy level.

Key equation: Photon energy equals the magnitude of the energy difference between the two levels.

Maximum Number of Spectral Lines

N = n(n − 1)/2

If an atom is excited to the nth energy level and all possible downward transitions are allowed, the maximum number of distinct transitions is n(n − 1)/2.

Example: For n = 4, the maximum number of transitions is:

N = 4 × 3/2 = 6

6. Solved Numerical Problems

Example 1: Energy of an Electron in the Second Orbit

Question: Calculate the energy of an electron in the second orbit of hydrogen.

Solution:

Using E_n = −13.6/n² eV:

E₂ = −13.6/2² = −3.4 eV

Answer: −3.4 eV

Example 2: Energy of an Emitted Photon

Question: Find the energy of the photon emitted when an electron in hydrogen falls from n = 3 to n = 2.

Solution:

E₃ = −13.6/9 = −1.51 eV approximately.

E₂ = −13.6/4 = −3.4 eV.

E_photon = E₃ − E₂ = 1.89 eV approximately

Answer: 1.89 eV

Example 3: Number of Spectral Lines

Question: What is the maximum number of spectral lines possible when an electron is excited to n = 5?

Solution:

N = 5(5 − 1)/2 = 10

Answer: 10 possible transitions

Example 4: Ionization Energy

Question: Calculate the energy required to ionize hydrogen from its ground state.

Solution:

The ground-state energy is −13.6 eV. Ionization takes the electron to E = 0.

E required = 0 − (−13.6) = 13.6 eV

Answer: 13.6 eV per atom

Example 5: Radius of the Third Orbit

Question: Calculate the radius of the third Bohr orbit of hydrogen.

Solution:

r₃ = 0.529 × 3² Å = 4.761 Å

Answer: 4.761 Å, or 4.761 × 10⁻¹⁰ m

7. JEE Main Preparation Tips

• Memorize the formulas for energy, radius, velocity and wavelength.
• Practice hydrogen-like ions such as He⁺ and Li²⁺.
• Check whether a question asks for emission or absorption.
• Convert electronvolts to joules when the question requires SI units: 1 eV ≈ 1.602 × 10⁻¹⁹ J.
• Remember that 1 Å = 10⁻¹⁰ m.
• For wavelength questions, identify both initial and final levels before substituting values.
• Revise NCERT theory and solve previous-year JEE Main questions.

8. Practice MCQs for JEE Main

Q1. The energy of an electron in the ground state of hydrogen is:

A. −3.4 eV
B. −13.6 eV
C. +13.6 eV
D. 0 eV

Show Answer

Correct option: B. −13.6 eV

Q2. The radius of the second Bohr orbit of hydrogen is how many times the radius of the first orbit?

A. 2
B. 3
C. 4
D. 8

Show Answer

Correct option: C. 4, because r is proportional to n².

Q3. The Lyman series of hydrogen lies mainly in which region?

A. Infrared
B. Ultraviolet
C. Microwave
D. Radio waves

Show Answer

Correct option: B. Ultraviolet

Q4. The energy of the photon emitted in the transition n = 2 to n = 1 in hydrogen is:

A. 3.4 eV
B. 10.2 eV
C. 13.6 eV
D. 1.89 eV

Show Answer

Correct option: B. 10.2 eV, since −3.4 − (−13.6) = 10.2 eV.

Q5. The maximum number of transitions from an atom initially excited to n = 4 is:

A. 4
B. 5
C. 6
D. 8

Show Answer

Correct option: C. 6

Q6. The ground-state ionization energy of He⁺ is:

A. 13.6 eV
B. 27.2 eV
C. 40.8 eV
D. 54.4 eV

Show Answer

Correct option: D. 54.4 eV, using 13.6 × Z² eV with Z = 2.

Q7. When an electron moves from n = 2 to n = 3 in hydrogen, the atom:

A. Emits a photon
B. Absorbs energy
C. Emits an alpha particle
D. Loses its proton

Show Answer

Correct option: B. Absorbs energy

9. Atomic Physics: Quick Revision Sheet

Revise These Before the Exam

• Bohr's angular momentum: mvr = nℏ.
• Energy: Eₙ = −13.6 Z²/n² eV.
• Radius: rₙ = 0.529 n²/Z Å.
• Velocity: vₙ = 2.18 × 10⁶ Z/n m/s.
• Photon energy: E = hν = hc/λ.
• Rydberg equation: 1/λ = RZ²(1/nf² − 1/ni²).
• Lyman: nf = 1; Balmer: nf = 2; Paschen: nf = 3.
• Maximum transitions: N = n(n − 1)/2.
• Hydrogen ground-state ionization energy: 13.6 eV.
• Hydrogen ground-state excitation to n = 2: 10.2 eV.

10. Frequently Asked Questions (FAQs)

Q1. Is Atomic Physics important for JEE Main?

Atomic structure and energy-level concepts are useful parts of the modern physics syllabus. Study them alongside the other topics listed in the current official JEE Main syllabus.

Q2. Which formulas should I memorize first?

Start with the Bohr energy and radius formulas, the Rydberg equation, photon energy, and the formula for the maximum number of transitions.

Q3. What is the difference between excitation and ionization?

Excitation moves an electron to a higher bound state. Ionization supplies enough energy to remove the electron completely from the atom.

Q4. Can Bohr's model explain every atom?

No. It works well for hydrogen-like, one-electron species, but it cannot fully explain the spectra of multi-electron atoms.

🎯 Keep Practicing and Keep Improving!

Understanding concepts is more effective than memorizing formulas without practice. Revise this guide, solve numerical problems, and test yourself with previous-year questions.

Share this post with your Class 12 and JEE Main study group.

Happy Learning! 📘