Dual Nature of Matter

Introduction

Just as light exhibits both wave and particle nature, Louis de Broglie proposed in 1924 that matter also has wave-like properties. This revolutionary idea earned him the Nobel Prize and laid the foundation for quantum mechanics.

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de Broglie Hypothesis

The Wave-Particle Duality of Matter

de Broglie suggested that every moving particle has an associated wavelength:

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

where:

• $\lambda$ = de Broglie wavelength
• $h$ = Planck’s constant = $6.626 \times 10^{-34}$ J·s
• $m$ = mass of particle
• $v$ = velocity of particle
• $p$ = momentum = $mv$

Interactive Demo: Visualize Matter Waves

See the wave nature of matter particles and how wavelength changes with momentum.

Key Insights

1. Wavelength is inversely proportional to momentum
2. Heavier objects have negligibly small wavelengths
3. Only significant for microscopic particles (electrons, protons)

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de Broglie Wavelength Formulas

For Particles with Given KE

If kinetic energy $KE = \frac{1}{2}mv^2$:

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

For Accelerated Charged Particles

For a particle with charge $q$ accelerated through potential $V$:

$$KE = qV$$ $$\boxed{\lambda = \frac{h}{\sqrt{2mqV}}}$$

For Electrons (Simplified)

For electrons accelerated through potential V (in volts):

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

This is a very useful formula for quick calculations!

Thermal de Broglie Wavelength

For particles at temperature T:

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

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Calculations

Example 1: Electron Wavelength

Problem: Find the de Broglie wavelength of an electron moving at $10^6$ m/s.

Solution:

$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6}$$ $$\lambda = 7.28 \times 10^{-10} \text{ m} = 7.28 \text{ Å}$$

Example 2: Accelerated Electron

Problem: An electron is accelerated through 100 V. Find its wavelength.

Solution: Using the simplified formula:

$$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ Å}$$

Example 3: Proton vs Electron

Problem: An electron and proton have the same kinetic energy. Compare their wavelengths.

Solution:

$$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e KE}}{h/\sqrt{2m_p KE}} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1836} \approx 42.8$$

The electron has ~43 times larger wavelength!

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Verification of Bohr’s Quantization

de Broglie showed that Bohr’s angular momentum quantization follows naturally if we assume an integral number of wavelengths fit in an orbit:

$$2\pi r = n\lambda$$

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

$$2\pi r = n \times \frac{h}{mv}$$ $$mvr = \frac{nh}{2\pi}$$

This is exactly Bohr’s quantization condition!

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Heisenberg Uncertainty Principle

Statement

It is impossible to simultaneously determine the exact position and momentum of a particle:

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

or equivalently:

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

where:

• $\Delta x$ = uncertainty in position
• $\Delta p$ = uncertainty in momentum
• $\Delta v$ = uncertainty in velocity

Energy-Time Form

There’s also an energy-time uncertainty relation:

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

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Implications of Uncertainty Principle

1. Electrons Cannot Exist in the Nucleus

If an electron were confined to the nucleus (Δx ≈ 10⁻¹⁴ m):

$$\Delta v \geq \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 10^{-14}}$$ $$\Delta v \geq 5.8 \times 10^9 \text{ m/s}$$

This exceeds the speed of light! So electrons can’t be in the nucleus.

2. Bohr’s Model is Fundamentally Flawed

Bohr’s model specifies exact orbits (definite r and v), violating the uncertainty principle.

Example Calculation

Problem: The uncertainty in position of an electron is 0.1 Å. Calculate the minimum uncertainty in its velocity.

Solution:

$$\Delta v = \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 10^{-11}}$$ $$\Delta v = 5.79 \times 10^{6} \text{ m/s}$$

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Comparison of Macroscopic vs Microscopic

Aspect	Macroscopic (Ball)	Microscopic (Electron)
de Broglie λ	~10⁻³⁴ m	~10⁻¹⁰ m
Wave effects	Unobservable	Significant
Uncertainty	Negligible	Fundamental

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Key Points for JEE

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Practice Problems

1. An α-particle and a proton are accelerated through the same potential. Find the ratio of their de Broglie wavelengths.

2. The uncertainty in position of a particle is equal to its de Broglie wavelength. What is the minimum uncertainty in its velocity?

3. Calculate the wavelength of a neutron with kinetic energy 0.025 eV (thermal neutron).

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