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Modern Physics

de Broglie Waves and Matter Waves

de Broglie waves represent the wave nature associated with every moving material particle, extending the concept of wave-particle duality from radiation to matter. The wavelength is inversely proportional to the linear momentum of the particle, linking particle mechanics with wave mechanics.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Qualitative understanding of matter waves, de Broglie wavelength formula, dependence on mass, velocity, and accelerating potential, and basic numerical calculations for electrons and gas particles.

PGT extension

Derivation of de Broglie wavelength from relativistic and non-relativistic mechanics, phase and group velocity relations, Heisenberg's uncertainty principle connection, and wave packet interpretation.

Prerequisites

Planck's quantum theory of radiation
Special theory of relativity (mass-energy equivalence)
Wave mechanics fundamentals (phase velocity, group velocity)

Concept of Matter Waves

Louis de Broglie proposed in 1924 that moving material particles exhibit wave-like properties under appropriate conditions.
The associated wavelength lambda is given by lambd$a = h/p$, where h is Planck's constant and p is the linear momentum.
Matter waves are not electromagnetic waves; they represent the probability amplitude of finding the particle in space.

Wavelength Expressions for Different Particles

For a charged particle of mass m and charge q accelerated through a potential difference V, lambd$a = h / \sqrt(2mqV)$.
For an electron, substituting fundamental constants yields lambd$a = 12.27 / \sqrt(V) Angstroms$.
For a gas molecule or thermal neutron at absolute temperature T, the kinetic energy is substituted as $E = (3/2)kT$, giving lambd$a = h / \sqrt(3mkT)$.

Wave-Particle Duality and Experimental Verification

Davisson-Germer experiment and G.P. Thomson experiment provided direct experimental confirmation of electron diffraction, verifying de Broglie's hypothesis.
Bragg's law (2d sin(theta) = n*lambda) is used in crystal diffraction experiments to measure the wavelength of matter waves.
Macroscopic objects possess negligibly small de Broglie wavelengths due to their large mass, making their wave nature undetectable.

Phase Velocity and Group Velocity

A particle is represented by a wave packet formed by the superposition of waves of slightly different wavelengths.
The group velocity (v_g) of the wave packet equals the classical particle velocity (v).
The phase velocity (v_p) of the de Broglie waves exceeds the speed of light in vacuum (c), given by $v_p = c^2 / v$ for a relativistic particle.

Derivations

Derivation of de Broglie Wavelength

Conditions: From Planck's quantum theory and Einstein's mass-energy relation

According to Planck, the energy of a photon of frequency nu is $E = h*\nu$.
According to Einstein's mass-energy equivalence, the equivalent mass is $E = mc^2$.
Equating the two expressions: h*n$u = mc^2$, or h*(c/lambda) = mc^2, which simplifies to lambd$a = h/(mc) = h/p$.

Result: lambd$a = h/p$

de Broglie Wavelength of an Accelerated Electron

Conditions: Non-relativistic charged particle accelerated through potential V

Kinetic energy acquired by an electron of charge e accelerated through potential V is E = eV.
Linear momentum p in terms of kinetic energy is p = sqrt(2mE) = sqrt(2meV).
Substituting p into the de Broglie relation gives lambd$a = h / \sqrt(2meV)$.
Substituting numerical values of h (6.626 x 10^-34 J s), m (9.109 x 10^-31 kg), and e (1.602 x 10^-19 C) yields lambd$a = 12.27 / \sqrt(V) Angstroms$.

Result: lambd$a = (12.27 / \sqrt(V)) \text{ Å}$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
General de Broglie Wavelength	$$\lambda = \frac{h}{p} = \frac{h}{mv}$$	h = Planck constant, p = momentum, m = mass, v = velocity m	Valid for all material particles
Wavelength in terms of Kinetic Energy	$$\lambda = \frac{h}{\sqrt{2mE}}$$	E = Kinetic energy of the particle m	Non-relativistic regime
Wavelength of Charged Particle in Potential V	$$\lambda = \frac{h}{\sqrt{2mqV}}$$	q = charge of particle, V = accelerating potential m	Non-relativistic charged particles
Electron Wavelength Formula	$$\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}$$	V = potential in volts \u00c5 (Angstroms)	Non-relativistic electrons
Thermal Neutron / Gas Molecule Wavelength	$$\lambda = \frac{h}{\sqrt{3mkT}}$$	k = Boltzmann constant, T = absolute temperature in Kelvin m	Thermal equilibrium at temperature T
Relativistic de Broglie Wavelength	$$\lambda = \frac{h}{\sqrt{2m_0 E + \frac{E^2}{c^2}}}$$	$m_0 = rest mass$, E = kinetic energy, c = speed of light m	High velocities approaching speed of light
Group Velocity of Matter Waves	$$v_g = \frac{d\omega}{dk} = v$$	omega = angular frequency, k = propagation constant, v = particle velocity m/s	Wave packet representing a particle
Phase Velocity of Matter Waves	$$v_p = \nu\lambda = \frac{E}{p} = \frac{c^2}{v}$$	nu = frequency, lambda = wavelength, v = particle velocity m/s	Relativistic particle waves

Graphs and experiments

Davisson-Germer Experiment

Draw: Polar graph of scattered electron intensity versus scattering angle theta for different accelerating voltages, showing a distinct secondary maximum at 50 degrees for 54 V.

Interpret: Demonstrates constructive interference of electron waves scattered from atomic planes of a nickel crystal, confirming the de Broglie wavelength relation.

Exam use: Tested via questions on accelerating potential, crystal lattice spacing, and Bragg diffraction angle.

Wavelength versus Potential (Lambda vs 1/sqrt(V))

Draw: A straight line passing through the origin representing lambda plotted against 1/sqrt(V).

Interpret: Shows inverse square-root dependence of de Broglie wavelength on accelerating potential for electrons.

Exam use: Numerical slope interpretation and graphical proportionality identification in TGT/PGT.

Wave Packet Dispersion

Draw: A localized wave envelope moving with group velocity v_g while internal phase ripples move at phase velocity v_p.

Interpret: Illustrates how a localized particle is constructed by superimposing waves of nearby frequencies.

Exam use: Concept testing on group velocity versus particle velocity in advanced PGT sections.

Solved numericals

TGT core

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.

Use the formula: lambd$a = 12.27 / \sqrt(V) Angstroms$.
Substitute V = 100 V into the formula.
lambd$a = 12.27 / \sqrt(100) = 12.27 / 10 = 1.227 Angstroms$.

Answer: 1.227 Angstroms (or 1.227 x 10^-10 m)

PGT extension

Find the de Broglie wavelength of a neutron of mass 1.67 x 10^-27 kg kept in thermal equilibrium at a temperature of 27 degree Celsius.

Convert temperature to Kelvin: $T = 27 + 273 = 300 K$.
Use the thermal neutron wavelength formula: lambd$a = h / \sqrt(3mkT)$.
Substitute values: $h = 6.63 x 10^-34 J s$, $m = 1.67 x 10^-27 kg$, $k = 1.38 x 10^-23 J/K$.
Denominato$r = \sqrt(3 * 1.67 x 10^-27 * 1.38 x 10^-23 * 300) = \sqrt(2.07 x 10^-46) = 1.44 x 10^-23$.
lambd$a = (6.63 x 10^-34) / (1.44 x 10^-23) = 4.60 x 10^-11 m$.

Answer: 0.460 nm (or 4.60 x 10^-11 m)

PGT extension

At what velocity is the de Broglie wavelength of an electron equal to its Compton wavelength?

Relativistic de Broglie wavelength is lambd$a = h / (\gamma * m_0 * v)$, where gamm$a = 1 / \sqrt(1 - v^2/c^2)$.
Compton wavelength is lambd$a_c = h / (m_0 * c)$.
Equating the two: h / (gamma * m_0 * v) = h / (m_0 * c), which gives gamma * v = c.
Substitute gamm$a = 1 / \sqrt(1 - v^2/c^2): (v / \sqrt(1 - v^2/c^2)) = c$.
Squaring both sides: v^2 / (1 - $v^2/c^2) = c^2 => v^2 = c^2 - v^2 => 2v^2 = c^2$.
Solving for v: $v = c / \sqrt(2)$.

Answer: $v = c / \sqrt(2) (or 0.707 c)$

Assumptions and limitations

The de Broglie hypothesis applies to both microscopic and macroscopic entities, but limitations arise because macroscopic objects have unmeasurably small wavelengths.
Non-relativistic formulas assume v << c; relativistic corrections must be incorporated for high-energy particles.
Matter waves are statistical probability waves and do not represent mechanical vibrations in a physical medium.

Common misconceptions

Common error: Matter waves are electromagnetic waves emitted by moving charged particles.

Correct: Matter waves are non-electromagnetic probability waves associated with any moving mass.

Electromagnetic waves are oscillations of electric and magnetic fields, whereas de Broglie waves describe the probability amplitude of particle position.

Common error: The phase velocity of matter waves is less than or equal to the speed of light in vacuum.

Correct: The phase velocity of matter waves always exceeds the speed of light ($v_p = c^2/v)$.

Since particle velocity v < c, c^2/v is strictly greater than c, but energy or information does not travel at phase velocity.

Common error: A stationary particle has a very large de Broglie wavelength approaching infinity.

Correct: A stationary particle in free space has an undefined or zero de Broglie wavelength because momentum is zero.

lambd$a = h/p implies that$ as p approaches zero, wavelength diverges mathematically, but a truly stationary free particle has no wave association.

Quick revision

de Broglie wavelength formula: lambd$a = h/p = h/(mv)$.
For electrons: lambd$a = 12.27 / \sqrt(V) Angstroms$.
Davisson-Germer experiment proved the wave nature of electrons via crystal diffraction.
Group velocity of matter waves equals particle velocity ($v_g = v)$, while phase velocity exceeds light speed ($v_p = c^2/v)$.

Must remember

The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
For an electron accelerated through a potential difference $V$, the wavelength is $\lambda = \frac{12.27}{\sqrt{V}} \text{ \u00c5}$.
The phase velocity of matter waves is $v_p = \frac{c^2}{v}$, which always exceeds the speed of light in vacuum.
The group velocity of matter waves equals the classical particle velocity ($v_g = v$).
Thermal neutrons or gas molecules at absolute temperature $T$ have a de Broglie wavelength of $\lambda = \frac{h}{\sqrt{3mkT}}$.

Glossary

Matter WavesWaves associated with moving material particles representing the probability amplitude of finding the particle.

Group VelocityThe velocity with which the wave packet (representing the particle) travels, equal to particle velocity $v$.

Phase VelocityThe velocity of propagation of a single phase component of the de Broglie wave, given by $v_p = \frac{c^2}{v}$.

General de Broglie Wavelength\lambda = \frac{h}{p} = \frac{h}{mv} (Valid for all material particles)

Phase Velocity vs Group Velocity

Feature	Phase Velocity ($v_p$)	Group Velocity ($v_g$)
Definition	Velocity of individual wave crests	Velocity of the wave packet envelope
Relation to $c$	Exceeds the speed of light ($v_p > c$)	Equals particle velocity ($v_g = v < c$)
Physical Meaning	Speed of constant phase propagation	Speed of particle and energy transport

Practice Quiz

Choose an option to reveal the explanation.

1. What is the de Broglie wavelength of a particle of mass $m$ and momentum $p$?

Explanation: According to de Broglie's hypothesis, the associated wavelength is $\lambda = \frac{h}{p}$.

2. For an electron accelerated through a potential difference of $V$ volts, the de Broglie wavelength in Angstroms is given by:

Explanation: Substituting fundamental constants for an electron yields $\lambda = \frac{12.27}{\sqrt{V}} \text{ \u00c5}$.

3. Which of the following statements regarding matter waves is correct?

Explanation: Matter waves are non-electromagnetic waves representing the probability amplitude of locating a particle in space.

4. The group velocity of de Broglie waves associated with a particle is equal to:

Explanation: The group velocity ($v_g$) of the wave packet representing a particle equals the classical particle velocity ($v$).

5. The phase velocity of a relativistic matter wave is given by:

Explanation: For a relativistic particle, the phase velocity is $v_p = \frac{c^2}{v}$, which exceeds the speed of light since $v < c$.

6. What is the de Broglie wavelength of a gas molecule of mass $m$ at absolute temperature $T$?

Explanation: Substituting the kinetic energy $E = \frac{3}{2}kT$ into $\lambda = \frac{h}{\sqrt{2mE}}$ yields $\lambda = \frac{h}{\sqrt{3mkT}}$.

7. Davisson-Germer experiment provided direct experimental verification of:

Explanation: The Davisson-Germer experiment confirmed electron diffraction, proving the wave nature of electrons proposed by de Broglie.

8. Why are wave properties not observed for macroscopic objects?

Explanation: Because Planck's constant $h$ is extremely small and macroscopic masses are large, $\lambda = \frac{h}{mv}$ yields an undetectable wavelength.

9. Bragg's law used in crystal diffraction for matter waves is expressed as:

Explanation: Bragg's law for constructive interference from crystal planes is $2d \sin(\theta) = n \lambda$.

10. If the kinetic energy of a non-relativistic particle is doubled, its de Broglie wavelength becomes:

Explanation: Since $\lambda \propto \frac{1}{\sqrt{E}}$, doubling $E$ decreases $\lambda$ by a factor of $\sqrt{2}$.

11. Which condition makes the relativistic formula for de Broglie wavelength necessary?

Explanation: At velocities approaching $c$, relativistic mass and energy corrections must be used for momentum and wavelength calculations.

12. The relation between frequency $\nu$ of a de Broglie wave and particle energy $E$ is given by:

Explanation: Planck's relation $E = h\nu$ connects the frequency of the matter wave to the total energy of the particle.

13. What is the nature of the phase velocity of matter waves?

Explanation: Phase velocity $v_p = \frac{c^2}{v} > c$, but since no energy or information travels at phase velocity, special relativity is not violated.

14. A proton and an $\alpha$ particle are accelerated through the same potential difference $V$. The ratio of their de Broglie wavelengths ($\lambda_p / \lambda_\alpha$) is:

Explanation: Formula: $\lambda = \frac{h}{\sqrt{2mqV}}$. Mass of $\alpha$ $m_\alpha = 4 m_p$, charge $q_\alpha = 2 q_p$. Thus $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = 2\sqrt{2}$.

15. A particle is described by a wave packet rather than a single monochromatic wave because:

Explanation: A single infinite sine wave has uncertain position; superposition into a wave packet localizes the particle in space.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. An electron and a photon have the same wavelength $\lambda$. The ratio of their momenta is:

Explanation: Momentum for both is given by $p = \frac{h}{\lambda}$. Since their wavelengths are equal, their momenta must be equal, giving a ratio of $1:1$.

2. The de Broglie wavelength of an electron accelerated through a potential of $100\text{ V}$ is approximately:

Explanation: Using $\lambda = \frac{12.27}{\sqrt{V}} = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ \u00c5}$.

3. If the momentum of a particle is increased by $100\%$, its de Broglie wavelength will change by:

Explanation: Since $\lambda = \frac{h}{p}$, if momentum becomes $2p$, the new wavelength is $\lambda' = \frac{\lambda}{2}$, which is a $50\%$ decrease.

4. The de Broglie wavelength of a thermal neutron at temperature $T$ is proportional to:

Explanation: From $\lambda = \frac{h}{\sqrt{3mkT}}$, wavelength is inversely proportional to the square root of absolute temperature ($T^{-1/2}$).

5. Which of the following relations correctly expresses the de Broglie wavelength in terms of kinetic energy $E$ for a non-relativistic particle?

Explanation: Since $p = \sqrt{2mE}$, substituting this into $\lambda = \frac{h}{p}$ gives $\lambda = \frac{h}{\sqrt{2mE}}$.

6. In the Davisson-Germer experiment, the scattered beam of electrons shows a peak intensity at a specific accelerating voltage and scattering angle. For $54\text{ V}$, the scattering angle is:

Explanation: In standard Davisson-Germer setup, peak intensity occurs at $V = 54\text{ V}$ and scattering angle $\theta = 50^\circ$ (or glancing angle $\alpha = 65^\circ$).

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the de Broglie wavelength of an electron with kinetic energy equal to $100\text{ eV}$?

Explanation: $\lambda = \frac{12.27}{\sqrt{100}} \text{ \u00c5} = 1.227 \text{ \u00c5} = 0.1227 \text{ nm} \approx 0.123 \text{ nm}$.

2. The phase velocity $v_p$ and group velocity $v_g$ of a matter wave satisfy which of the following relations?

Explanation: Since $v_g = v$ and $v_p = \frac{c^2}{v}$, multiplying them gives $v_p v_g = \left(\frac{c^2}{v}\right)(v) = c^2$.

3. If a deuteron and an $\alpha$ particle are accelerated through the same potential difference, the ratio of their de Broglie wavelengths is:

Explanation: Deuteron: $m_d = 2m_p, q_d = q_p$. $\alpha$: $m_\alpha = 4m_p, q_\alpha = 2q_p$. Ratio $\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_d q_d}} = \sqrt{\frac{4 \times 2}{2 \times 1}} = \sqrt{4} = 2:1$.

4. A particle of mass $m$ has kinetic energy $E$. The de Broglie wavelength expressed relativistically is:

Explanation: The relativistic de Broglie wavelength formula incorporates total energy / rest mass energy relation resulting in $\lambda = \frac{h}{\sqrt{2m_0 E + \frac{E^2}{c^2}}}$.

5. The concept of matter waves was proposed by Louis de Broglie in which year?

Explanation: Louis de Broglie proposed the hypothesis of matter waves in his doctoral thesis in 1924.

6. What happens to the de Broglie wavelength of a moving particle if its velocity is increased?

Explanation: Since $\lambda = \frac{h}{mv}$, an increase in velocity $v$ results in a decrease in wavelength $\lambda$.

7. The de Broglie wavelength of an uncharged particle (neutron) of mass $m$ at temperature $T$ is:

Explanation: Thermal neutrons have kinetic energy $E = \frac{3}{2}kT$, yielding $\lambda = \frac{h}{\sqrt{3mkT}}$.

8. Which experiment confirmed the wave nature of electrons independently of Davisson and Germer?

Explanation: G.P. Thomson performed transmission electron diffraction experiments on thin metal foils, confirming de Broglie's wave hypothesis.

9. A stationary particle in free space has a de Broglie wavelength that is:

Explanation: A truly stationary free particle has zero momentum, making $\lambda = \frac{h}{0}$ mathematically undefined / non-existent as a wave packet.

10. If the kinetic energy of a free electron is increased by a factor of 4, the percentage change in its de Broglie wavelength is:

Explanation: New wavelength $\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}$ (a $50\%$ reduction in value, meaning it becomes half, so percentage change in magnitude is a decrease to $50\%$, hence a $50\%$ decrease from initial... wait, $\lambda' = 0.5\lambda$ means a $50\%$ decrease).