JEE Physics: Modern Physics Complete Guide

Modern Physics is a crucial section in JEE Physics, encompassing the study of atomic and nuclear phenomena, quantum mechanics, photoelectric effect, and much more. This guide provides an in-depth overview of the topics, formulas, and concepts needed to excel in this area. Understanding these topics will not only help you score well but also build a strong foundation for higher studies in physics and engineering.

1. Atomic Models and Structure

Atomic models have evolved to explain the structure and behavior of atoms:

Bohr’s postulates:

  1. Electrons move in certain allowed circular orbits without radiating energy.
  2. Angular momentum of electron is quantized: $$mvr = n\hbar = n\frac{h}{2\pi}$$ where \(n=1,2,3,...\)
  3. Energy is emitted or absorbed when electron jumps between orbits, given by: $$\Delta E = h\nu$$

Bohr Radius

The radius of the first orbit in hydrogen atom:

$$a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m e^2} \approx 0.529 \times 10^{-10} \text{m}$$

Energy Levels in Hydrogen Atom

Energy of \(n^{th}\) orbit is:

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

Negative sign indicates bound state. Ionization energy corresponds to \(n \to \infty\).

2. Photoelectric Effect

The photoelectric effect demonstrates the particle nature of light where electrons are ejected from a metal surface when light of sufficient frequency shines on it.

Einstein’s Photoelectric Equation

The maximum kinetic energy of emitted electrons:

$$K_{max} = h\nu - \phi$$

Where:
\(h\) = Planck's constant \(6.626 \times 10^{-34} \text{Js}\)
\(\nu\) = frequency of incident light
\(\phi\) = work function of metal (minimum energy to eject electron)

Important Points

3. Dual Nature of Matter

Electrons and other particles exhibit wave-particle duality.

De Broglie Wavelength

All matter has an associated wavelength given by:

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

This was experimentally confirmed by electron diffraction experiments.

4. X-Rays

X-rays are electromagnetic waves with high energy, produced by the deceleration of high-speed electrons (bremsstrahlung) or characteristic X-rays from atomic transitions.

Duane-Hunt Law

The shortest wavelength in the X-ray spectrum:

$$\lambda_{min} = \frac{hc}{eV}$$

Where \(V\) is the accelerating voltage.

5. Nuclear Physics

Deals with the structure, stability, and reactions of atomic nuclei.

Basic Terms

Mass-Energy Relation and Binding Energy

According to Einstein’s relation:

$$E = mc^2$$

The mass defect is the difference between mass of individual nucleons and nucleus. Binding energy:

$$BE = \text{(Mass defect)} \times c^2$$

Binding energy per nucleon indicates stability.

Radioactivity

Spontaneous nuclear decay emits alpha, beta, or gamma radiation.

Radioactive decay law:

$$N = N_0 e^{-\lambda t}$$

Where \(N_0\) is initial nuclei, \(\lambda\) is decay constant.

Half-life \(T_{1/2}\) is related to \(\lambda\):

$$T_{1/2} = \frac{\ln 2}{\lambda}$$

Types of Radioactive Decay

6. Nuclear Fission and Fusion

Nuclear Fission

Heavy nucleus splits into lighter nuclei releasing large energy, used in nuclear reactors.

Example:

$$^{235}U + n \to ^{92}Kr + ^{141}Ba + 3n + \text{energy}$$

Nuclear Fusion

Light nuclei combine to form heavier nuclei releasing energy, powering stars including the sun.

Example:

$$^2H + ^3H \to ^4He + n + \text{energy}$$

7. Important Constants and Formulas

Constant Value Remarks
Planck’s constant, \(h\) \(6.626 \times 10^{-34} \, \text{Js}\) Quantum of action
Speed of light, \(c\) \(3 \times 10^{8} \, \text{m/s}\) Electromagnetic wave speed
Electron charge, \(e\) \(1.6 \times 10^{-19} \, \text{C}\) Elementary charge
Mass of electron, \(m_e\) \(9.11 \times 10^{-31} \, \text{kg}\) Rest mass of electron
Bohr radius, \(a_0\) \(0.529 \times 10^{-10} \, \text{m}\) Radius of hydrogen atom’s 1st orbit

8. Practice Problems

Problem 1: Calculate the wavelength of an electron moving with velocity \(2 \times 10^6 \, \text{m/s}\).

Solution:

$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2 \times 10^{6}} = 3.64 \times 10^{-10} \text{m}$$

Problem 2: A metal has a work function of 2 eV. Calculate the stopping potential if incident light wavelength is 400 nm.

Solution:

Energy of photon,

$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \text{J} = 3.1 \, eV$$

Maximum kinetic energy:

$$K_{max} = 3.1 - 2 = 1.1 \, eV$$

Stopping potential,

$$V_0 = \frac{K_{max}}{e} = 1.1 \, \text{Volts}$$

9. Tips for JEE Exam

This comprehensive guide equips you with all the fundamental and advanced concepts in Modern Physics essential for JEE preparation. Consistent study and problem-solving practice on these topics will ensure a strong performance in the exam.