de Broglie’s Relationship

1.0Introduction to de Broglie’s Relationship

The de Broglie relationship is a fundamental concept in quantum mechanics which links the wave nature of particles with their momentum. Louis de Broglie, in 1924, proposed that matter exhibits both particle and wave characteristics.

This relationship established that every moving particle can be associated with a wavelength, now called the de Broglie wavelength, which is essential to understanding atomic structure, electron behaviour, and quantum mechanics.

2.0Historical Background

Before de Broglie, physics treated light as a wave (interference, diffraction) and particles (photoelectric effect).

• Einstein introduced the concept of photons to explain the photoelectric effect.
• Inspired by this, de Broglie hypothesized that particles, such as electrons, should also show wave-like properties.

This was a revolutionary idea because it extended the wave-particle duality from light to all matter.

3.0Concept of Wave-Particle Duality

• Wave Nature: Particles like electrons can produce interference and diffraction patterns, typical of waves.
• Particle Nature: Particles like electrons have mass and momentum, which define their particle properties.

The de Broglie relationship mathematically bridges these two aspects by connecting momentum (particle property)with wavelength (wave property).

4.0de Broglie Hypothesis

Louis de Broglie proposed that:

“Every particle of matter moving with momentum p has an associated wavelength λ, given by λ=hp, where h is Planck’s constant.”

• This hypothesis implies that not only photons, but electrons, protons, neutrons, and even atoms exhibit wave behavior.

Significance:

• Explains atomic spectra and electron behavior in atoms.
• Forms the basis for quantum mechanics and Schrödinger’s wave equation.

5.0Mathematical Formulation of de Broglie Wavelength

The de Broglie wavelength (λ) of a particle is given by:

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

Where:

• h = Planck’s constant
• p = momentum of the particle (p=mv)

For a particle of mass m moving with velocity v:

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

This shows that the wavelength is inversely proportional to the momentum:

• Higher momentum → smaller wavelength
• Lower momentum → larger wavelength

6.0Relation Between de Broglie Wavelength and Momentum

• Momentum (p) is the product of mass (m) and velocity (v).
• Therefore:

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

• For light particles like electrons, the de Broglie wavelength is significant and observable.
• For macroscopic objects (like a ball or car), the de Broglie wavelength is extremely small and not measurable.

Example Calculation:

• An electron with mass $9.11\times 10^{-31}\text{ kg}$ moving at $1\times 10^6\text{ m/s}$:

$\lambda =\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 10^6}\approx 7.27\times 10^{-10}\text{m }(0.727\text{nm})$

This wavelength is comparable to atomic spacing, explaining electron diffraction.

7.0de Broglie Wavelength for Electrons

For electrons accelerated through a potential difference V, their kinetic energy is:

$\frac{1}{2}m{v}^2=eV$

Where:

• e = charge of electron
• V = accelerating voltage

Velocity of electron:

$v=\sqrt{\frac{2eV}{m}}$

Hence, de Broglie wavelength of an electron:

$\lambda =\frac{h}{\sqrt{2meV}}$

• This formula is widely used in electron diffraction experiments.

8.0Experimental Verification

Davisson-Germer Experiment

• Conducted in 1927, electrons were fired on a nickel crystal.
• Observed diffraction patterns, confirming that electrons behave as waves.
• Verified the de Broglie hypothesis quantitatively.

Electron Diffraction

• Electrons passing through a thin crystalline film produce interference patterns like X-rays.
• Shows wave property of electrons, supporting quantum mechanics.

9.0Applications of de Broglie Relationship

1. Electron Microscopes:
   
   Use electrons instead of light for imaging.
   
   Shorter wavelength → higher resolution.
2. Understanding Atomic Structure:
   
   Explains Bohr’s quantization of electron orbits.
   
   Electron waveforms correspond to standing waves in atoms.
3. X-ray Diffraction Analogy:
   
   Electron diffraction allows crystal lattice analysis.
4. Quantum Mechanics Development:
   
   Forms the basis for Schrödinger’s equation and wave mechanics.
5. Nanotechnology:
   
   Electron wavelengths used in nano-scale imaging and devices.

10.0Important Formulas

1. de Broglie Wavelength:

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

2. Electron accelerated through voltage V:

$\lambda =\frac{h}{\sqrt{2meV}}$

3. Momentum of particle:

$p=mv$

4. Kinetic Energy of a particle

$\text{KE}=\frac{1}{2}m{v}^2$

11.0Solved  Problems

Problem

A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of an electron is 2:1, the mass of the particle is:- 

Solution:

Using the de Broglie relation 

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{m_e{v}_e}{m_p{v}_p}$

Given:

• v_p = 4v_e
• λ_p  =  2λ_e

Substitute:

$2=\frac{me}{4mp}$

Solving:

$m_p=\frac{m_e}{8}$

Answer: The mass of the particle = 18\frac{1}{8}81​ the mass of an electron.

Problem

If the velocity of the electron is 1.6×106 m s−11.6\times10^{6}\ \text{m s}^{-1}1.6×106 m s−1. The de Broglie wavelength associated with this electron is:

Solution 

Use de Broglie relation: λ= hmv 

Given:

v=1.6×10⁶ m s⁻¹

Planck’s constant h=6.626×10⁻³⁴ 

Electron mass m_e=9.109×10−³¹ kg

Compute:

$\lambda =\frac{6.626\times 10^{-34}}{(9.109\times 10^{-31})\times (1.6\times 10^6)}=\frac{6.626\times 10^{-34}}{\left(1.45744\times 10^{-24}\right)}\approx 4.547\times 10^{-10}\text{m}$

Convert to picometres (1 pm = 10⁻¹²):

Answer: 455 pm.