Matter Waves

1.0What are Matter Waves?

In the early 20th century, physics underwent a revolution. Building upon Albert Einstein's work on the photoelectric effect, which showed that light has a particle-like nature (photons), Louis de Broglie proposed a bold hypothesis in 1924. He suggested that matter, which we traditionally think of as particles (like electrons, protons, and atoms), also possesses wave-like properties. These waves associated with moving particles are known as matter waves or de Broglie waves. This concept introduced the idea of wave-particle duality to matter.

The de Broglie hypothesis marked a pivotal moment in the development of quantum mechanics, challenging the classical view that matter and energy are distinct. Instead, it proposed that they are two facets of the same fundamental entity.

2.0Historical Background – Wave-Particle Duality

Before De Broglie’s discovery, scientists already knew that light exhibits both wave and particle nature.

• The wave nature of light was proved by phenomena like interference and diffraction.
• Particle nature of light was explained by Einstein’s photoelectric effect, where light behaves like photons with quantized energy $(E=h\nu )$.

Louis de Broglie extended this wave-particle duality concept to all matter, suggesting that particles such as electrons also exhibit wave-like behavior.

This marked the beginning of quantum theory of matter, changing our understanding of physics forever.

3.0De Broglie Wavelength: The Core Concept

The central equation of de Broglie's theory relates a particle's momentum to its associated wavelength. The de Broglie wavelength (\lambda) is given by:

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

Where:

• h is Planck's constant = $\left(6.626\times 10^{-34}\mathrm{J}\mathrm{s}\right)$
• p is the momentum of the particle
• m is the mass of the particle
• v is the velocity of the particle

This formula shows a crucial inverse relationship: the greater a particle's momentum, the shorter its de Broglie wavelength. For everyday objects like a cricket ball, the mass and velocity are so large that the wavelength is infinitesimally small and impossible to detect. However, for microscopic particles like electrons or protons, the momentum is small, and the wavelength is of a measurable size, leading to observable wave-like behavior such as diffraction and interference.

Wavelength of a Charged Particle

For a charged particle (like an electron with charge e) accelerated through a potential difference V, its kinetic energy (KE) is given by eV.

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

From this, we can express momentum p as:

$p=\sqrt{2m(KE)}=\sqrt{2mev}$

Substituting this into the de Broglie wavelength equation gives:

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

For an electron, this becomes particularly important. By substituting the values of h, me​ (mass of electron), and e, we get a simplified formula:

${\lambda }_{\text{electron }}=\frac{12.27}{\sqrt{V}}\mathring{A}$

This equation is a crucial tool for solving numerical problems in JEE Physics.

4.0Derivation of De Broglie Wavelength

Let’s understand how De Broglie derived his famous equation.

Step 1: Energy of a photon (from Einstein)

E=hν

And since light has a momentum $p=\frac{E}{c},$

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

Step 2: De Broglie’s analogy

De Broglie suggested that this relation should also hold true for material particles.
So, for a moving particle of mass mm and velocity vv:

p=mv

Hence, the De Broglie wavelength is:

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

This is valid for non-relativistic speeds $(i.e.,\mathrm{v}\ll \mathrm{c})$.
For relativistic particles, momentum $\mathrm{p}=\gamma \mathrm{mv}$, so:

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

where, $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Davisson-Germer Experiment: Experimental Evidence

The Davisson-Germer experiment in 1927 provided the first definitive experimental confirmation of the de Broglie hypothesis. They aimed a beam of electrons at a nickel crystal. The crystal's atomic spacing acted as a diffraction grating for the electron waves.

They observed that the electrons were scattered in specific directions, forming a diffraction pattern of concentric rings, similar to what's seen with X-rays.

The results matched the predictions of Bragg's law for wave diffraction, confirming that electrons, previously thought of only as particles, were indeed exhibiting wave-like behavior. This experiment solidified the concept of wave-particle duality and earned Clinton Davisson and George Thomson the Nobel Prize in Physics.

5.0Experimental Verification of Matter Waves

The Davisson-Germer Experiment (1927) provided the first experimental proof of the wave nature of electrons, validating De Broglie’s hypothesis.

Davisson-Germer Experiment Summary:

• Setup: A beam of electrons was accelerated and made to fall on a nickel crystal.
• Observation: The scattered electrons showed diffraction patterns, similar to X-ray diffraction.
• Conclusion: The wavelength calculated from the diffraction pattern matched the De Broglie wavelength.

This confirmed that electrons behave like waves under suitable conditions.

The De Broglie wavelength for an electron accelerated through a potential ( V ) is:

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

where:

• ( e ) = Electron charge
• ( m ) = Electron mass
• ( V ) = Accelerating potential

6.0Characteristics of Matter Waves

1. They are not electromagnetic in nature.
   Matter waves arise due to the motion of particles, not due to oscillating electric and magnetic fields.
2. Depends on the particle’s velocity.
   Faster particles have shorter wavelengths, and slower ones have longer wavelengths.
3. Invisible to the naked eye.
   The De Broglie wavelength for macroscopic bodies (like a cricket ball) is extremely small, hence negligible.
4. Probabilistic in interpretation.
   Matter waves represent the probability amplitude of finding a particle at a given position.
5. Confirmed experimentally through electron diffraction and neutron diffraction experiments.

7.0Relation Between Energy and Momentum

For a non-relativistic particle:

$E=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

Combining this with De Broglie’s relation $\lambda =\frac{h}{p}$;

$E=\frac{h^2}{2m{\lambda }^2}$

For a photon (relativistic case):

$E=pc=\frac{hc}{\lambda }$

Thus, the energy of matter waves depends inversely on the square of the wavelength for particles.

8.0Applications of Matter Waves

Understanding and applying matter waves has revolutionized modern science and technology.

1. Electron Microscopy

• Electron microscopes use electron beams with very short De Broglie wavelengths.
• They provide much higher resolution than optical microscopes.

2. Quantum Mechanics

• The Schrödinger wave equation is based on the concept of matter waves, representing the wavefunction of a particle.

3. Diffraction Studies

• Electron and neutron diffraction are used to determine atomic and molecular structures.

4. Semiconductor Technology

• Quantum tunneling and the behavior of electrons in potential wells rely on the wave nature of particles.

5. Nuclear and Particle Physics

• Matter waves explain the behavior of subatomic particles in accelerators and nuclear reactions.

9.0Key Properties of Matter Waves

1. They are not electromagnetic waves: Unlike light waves, matter waves are not associated with electric and magnetic fields. They are probability waves in a sense, where the amplitude of the wave at any point is related to the probability of finding the particle at that location.
2. Velocity is not constant: The velocity of a matter wave is not the speed of light (c). It can be greater or smaller than c, and it varies with the particle's velocity.
3. Independent of charge: The existence of a matter wave is independent of a particle's charge. Neutral particles like neutrons also exhibit wave-like properties.
4. Short-lived: Matter waves only exist as long as the particle is in motion.

10.0Solved Examples

1. Find the de Broglie wavelength of an electron moving at $v=3.0\times 10^6\mathrm{m}/\mathrm{s}$ . (Non-relativistic approximation)

Solution:
Non-relativistic de Broglie: $\lambda =\frac{h}{mv}$.

Plug in values:

$\lambda =\frac{6.626\times 10^{-34}}{\left(9.109\times 10^{-31}\right)\left(3.0\times 10^6\right)}$

Compute denominator: $9.109\times 10^{-31}\times 3.0\times 10^6=2.7327\times 10^{-24}$

So

$\lambda =\frac{6.626\times 10^{-34}}{2.7327\times 10^{-24}}\approx 2.426\times 10^{-10}\mathrm{m}.$

Answer: $2.43\times 10^{-10}\mathrm{m}(0.243\mathrm{nm})$

2. An electron is accelerated through a potential difference V=150 V. Calculate its de Broglie wavelength.

Solution:
Kinetic energy gained =eV. For non-relativistic electron:

$\lambda =\frac{h}{\sqrt{2m_{e}eV}}$

Plug numbers:

$\lambda =\frac{6.626\times 10^{-34}}{\sqrt{2\left(9.109\times 10^{-31}\right)\left(1.602\times 10^{-19}\right)(150)}}$

Compute the value (calculation shown):

$\lambda \approx 1.0014\times 10^{-10}\mathrm{m}$

Answer: $1.00\times 10^{-10}\mathrm{m}(0.100\mathrm{nm})$

3. Electrons accelerated through 150 V are incident on crystal planes of spacing d=0.25 nm. For first order Bragg reflection (n=1), find the Bragg angle $\theta$ (angle between plane and beam such that $(2\mathrm{d}\sin \theta =\mathrm{n}\lambda )$.

Solution:
From Problem 2, $\lambda \approx 1.00\times 10^{-10}\mathrm{m}$

Bragg law:

$n\lambda =2d\sin \theta \Rightarrow \sin \theta =\frac{n\lambda }{2d}$

Put values:

$\sin \theta =\frac{1.00\times 10^{-10}}{2\times 0.25\times 10^{-9}}=\frac{1.00\times 10^{-10}}{5.0\times 10^{-10}}=0.20$

$\text{ So }\theta =\arcsin (0.20)\approx 11.54^{\circ }\text{. }$

Answer: $\theta \approx 11.5^{\circ }$

4. The position of a particle is known to within $\Delta x=2.0\times 10^{-10}\mathrm{m}$. Find the minimum uncertainty in its momentum $\Delta \mathrm{p}$ and corresponding minimum uncertainty in velocity for an electron.

Solution:
Heisenberg:

$\Delta x\Delta p\ge \frac{\hslash }{2}$, $\text{ where }\hslash =\frac{h}{2\pi }$

So minimum $\Delta p=\frac{\hslash }{2\Delta x}$

Compute:

$\hslash =\frac{6.626\times 10^{-34}}{2\pi }=1.055\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$

Then

$\Delta p_{\min }=\frac{1.055\times 10^{-34}}{2\times 2.0\times 10^{-10}}=\frac{1.055\times 10^{-34}}{4.0\times 10^{-10}}=2.6375\times 10^{-25}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$

Corresponding $\Delta v=\Delta p/m_e$:

$\Delta v=\frac{2.6375\times 10^{-25}}{9.109\times 10^{-31}}\approx 2.896\times 10^5\mathrm{m}/\mathrm{s}$

Answer: $\Delta p_{\min }\approx 2.64\times 10^{-25}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s},\Delta v\approx 2.90\times 10^5\mathrm{m}/\mathrm{s}$

5. A baseball of mass 0.15 kg is moving at 40 m/s. Compute its de Broglie wavelength and comment why matter waves aren't observed for macroscopic objects.

Solution:

$\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}}{0.15\times 40}=\frac{6.626\times 10^{-34}}{6.0}\times 10^{-0}\approx 1.104\times 10^{-34}\mathrm{m}$

More direct calculation: Denominator

$0.15\times 40=6.0,\text{ so }\lambda =6.626\times 10^{-34}/6.0=1.1043\times 10^{-34}\mathrm{m}\text{. }$

Answer: $\lambda \approx 1.10\times 10^{-34}\mathrm{m}$​ — utterly negligible; far smaller than atomic nucleus scales, so no observable wave behavior.

6. An electron has kinetic energy $K=1.00MeV$. Compute its de Broglie wavelength (relativistic treatment).

Solution:

$\text{ Relativistic total energy }E=K+m_e{c}^2\text{, with }{m}_e{c}^2\approx 0.511\mathrm{MeV}\text{. }$

Work in joules:

$\text{ Convert: }1\mathrm{eV}=1.602\times 10^{-19}\mathrm{J}\text{. }$

So, $K=1.00\times 10^6\times 1.602\times 10^{-19}=1.602\times 10^{-13}\mathrm{J}\text{. }$

$m_e{c}^2=0.511\mathrm{MeV}=0.511\times 10^6\times 1.602\times 10^{-19}=8.187\times 10^{-14}\mathrm{J}$$\text{ Total energy }E=1.602\times 10^{-13}+8.187\times 10^{-14}=2.4207\times 10^{-13}\mathrm{J}\text{. }$

Momentum: $p=\frac{\sqrt{E^2-{\left(m_e{c}^2\right)}^2}}{c}$

Compute numerator inside sqrt:

$E^2-{\left(m_e{c}^2\right)}^2={\left(2.4207\times 10^{-13}\right)}^2-{\left(8.187\times 10^{-14}\right)}^2$

(Performing the arithmetic gives $p\approx 7.60\times 10^{-22}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}.)$

Then $\lambda =\frac{h}{p}\approx 8.72\times 10^{-13}\mathrm{m}$.

(Intermediate numeric work omitted here — final numeric value shown.)

Answer: $\lambda \approx 8.72\times 10^{-13}\mathrm{m}(0.872\mathrm{pm})$

7. Electrons with de Broglie wavelength $\lambda =1.00\times 10^{-10}\mathrm{m}$ pass through a double slit separated by $d=5.0\times 10^{-6}\mathrm{m}$. A screen is placed at distance $L=1.0$. Find the fringe spacing $\Delta \mathrm{y}$ on the screen (small angle approximation).

Solution:
Double slit fringe spacing $\Delta y=\frac{\lambda L}{d}$

$\Delta y=\frac{1.00\times 10^{-10}\times 1.0}{5.0\times 10^{-6}}=\frac{1.00\times 10^{-10}}{5.0\times 10^{-6}}=2.00\times 10^{-5}\mathrm{m}$

Answer: $\Delta y=2.0\times 10^{-5}\mathrm{m}=20\mu \mathrm{m}$

8. An electron is confined in a one-dimensional infinite potential well (length L=0.5 nm). For the ground state, show that the wavelength of the associated matter wave equals 2L and compute the de Broglie wavelength and momentum for the ground state.

Solution:
For infinite well, stationary wavefunction for ground state n=1 has half-wavelength fitting the box:${\lambda }_{\text{wave }}=2L$ . This is because $\psi$ has one half-wave in $0\to \mathrm{L}$.

So $\lambda =2L=2\times 0.5\times 10^{-9}=1.0\times 10^{-9}\mathrm{m}$

Momentum magnitude for a free-particle plane wave $p=\frac{h}{\lambda }$:

$p=\frac{6.626\times 10^{-34}}{1.0\times 10^{-9}}=6.626\times 10^{-25}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$

Kinetic energy $E=\frac{p^2}{2m}:$

$E=\frac{{\left(6.626\times 10^{-25}\right)}^2}{2\times 9.109\times 10^{-31}}\approx \frac{4.390\times 10^{-49}}{1.8218\times 10^{-30}}\approx 2.410\times 10^{-19}\mathrm{J}$ .

$\text{ Convert to }\mathrm{eV}:2.410\times 10^{-19}/1.602\times 10^{-19}\approx 1.50\mathrm{eV}\text{. }$

Answer: $\lambda =1.0\times 10^{-9}\mathrm{m},p=6.63\times 10^{-25}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s},E\approx 1.50\mathrm{eV}$