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Modern Physics
Wave-Particle Duality and De Broglie Hypothesis
Wave-particle duality posits that all material particles and electromagnetic radiation exhibit both wave and particle characteristics depending on the experimental context. The de Broglie hypothesis quantitatively links the momentum of a particle to its associated wavelength.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
De Broglie wavelength formula, momentum and kinetic energy dependence, basic Davisson-Germer experiment concepts, and photoelectric effect fundamentals.
PGT extension
Relativistic de Broglie wavelength, phase and group velocities of matter waves, Heisenberg's uncertainty principle derivation from wave packets, and fine details of electron diffraction.
Prerequisites
Planck's quantum theory of radiation
Special theory of relativity (energy-momentum relations)
Classical mechanics of waves and particles
De Broglie Hypothesis of Matter Waves
Louis de Broglie proposed in 1924 that moving material particles possess an associated wave nature, just like radiation.
The wavelength lambda associated with a particle of mass m moving with velocity v is given by lambd$a = h / (mv)$, where h is Planck's constant.
Matter waves are probability waves and do not require a physical medium for propagation.
Davisson-Germer Experiment
First experimental confirmation of the wave nature of electrons through electron diffraction from a nickel single crystal.
Observed a sharp maximum in scattered electron intensity at an accelerating potential of 54 V and scattering angle of 50 degrees.
The measured wavelength matched precisely with the theoretical de Broglie wavelength calculated for 54 eV electrons.
Phase Velocity and Group Velocity
A matter wave is represented as a wave packet rather than a single monochromatic wave, requiring the distinction between phase velocity (vp) and group velocity (vg).
The group velocity of the matter wave packet is equal to the classical particle velocity v (vg = v).
The phase velocity of matter waves exceeds the speed of light in vacuum (vp = c2 / v > c), but carries no physical signal.
Heisenberg's Uncertainty Principle as a Consequence
Wave-particle duality imposes an intrinsic limitation on the simultaneous measurement of conjugate variables.
The position-momentum uncertainty relation is given by Delta x Delta p >= h / (4 pi).
The time-energy uncertainty relation is given by Delta E Delta t >= h / (4 pi).
Derivations
De Broglie Wavelength in Terms of Kinetic Energy and Accelerating Potential
Conditions: Non-relativistic charged particle of mass m and charge q accelerated through potential V
Start with the definition of de Broglie wavelength: lambd$a = h / p$.
Express momentum p in terms of kinetic energy K: p = \sqrt{2mK}.
Substitute kinetic energy for a charged particle accelerated through potential V: K = qV, hence p = \sqrt{2mqV}.
Substitute p back into the wavelength equation to obtain lambd$a = h / \sqrt{2mqV}$.
Result: lambda = \frac{h}{\sqrt{2mqV}}
Relativistic De Broglie Wavelength
Conditions: Particle moving at velocities comparable to the speed of light c
Use the relativistic total energy relation E2 = p2c2 + m02c4 and rest energy E0 = m0c2.
lambda = wavelength, h = Planck's constant, p = momentum, m = mass, v = velocity m
General validity for any moving particle
De Broglie Wavelength for Electron (Potential Form)
$$\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}$$
V = accelerating potential in volts Angstroms
Non-relativistic electrons
De Broglie Wavelength in Terms of Temperature
$$\lambda = \frac{h}{\sqrt{3mk_B T}}$$
$k_B = Boltzmann constant$, T = absolute temperature, m = mass of gas molecule m
Thermal neutrons or gas molecules in thermal equilibrium
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
$$v_p = \frac{\omega}{k} = \frac{c^2}{v}$$
c = speed of light in vacuum, v = particle velocity m/s
Matter waves in free space
Heisenberg Position-Momentum Uncertainty
$$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$$
Delta x = position uncertainty, Delta p = momentum uncertainty J s
Simultaneous measurement of position and momentum
Heisenberg Time-Energy Uncertainty
$$\Delta E \cdot \Delta t \ge \frac{h}{4\pi}$$
Delta E = energy uncertainty, Delta t = time uncertainty J s
Simultaneous measurement of energy and time
Bragg's Law for Electron Diffraction
$$2d \sin\theta = n\lambda$$
d = interplanar spacing, theta = glancing angle, n = diffraction order, lambda = wavelength m
Constructive interference from crystal planes
Graphs and experiments
Davisson-Germer Experiment Polar Plot
Draw: Polar graph of scattered electron intensity versus scattering angle theta at a fixed accelerating potential (e.g., 54 V), showing a hump or spike at 50 degrees.
Interpret: The hump confirms constructive interference of electron waves scattered from different atomic planes of the Ni crystal, verifying wave behavior.
Exam use: Directly tested in conceptual MCQs regarding diffraction peak angles and accelerating voltages.
De Broglie Wavelength vs 1 / \sqrt{V} Plot
Draw: A straight line passing through the origin on a log-log or linear plot of lambda versus 1 / \sqrt{V}.
Interpret: Confirms the inverse square-root dependence of de Broglie wavelength on accelerating potential for charged particles.
Exam use: Interpreting slope to determine Planck's constant or charge-to-mass ratio.
Wave Packet Envelope (Dispersion)
Draw: An amplitude-modulated wave packet showing a localized group of waves moving with velocity vg while individual phase ripples move at vp.
Interpret: Illustrates how a localized particle is constructed from a superposition of monochromatic matter waves with a narrow band of wave numbers.
Exam use: Conceptual questions distinguishing phase velocity from group velocity in advanced PGT exams.
Solved numericals
TGT core
Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.
Use the simplified formula for electrons: lambd$a = 12.27 / \sqrt{V} Angstroms$.
Calculate lambd$a = h / p = 6.63 \times 10^{-34} / (2.31 \times 10^{-20}) = 2.87 \times 10^{-14} \text{ m}$.
Answer: 2.87 \times 10^{-14} \text{ m}
PGT extension
An electron has a speed of 1.8 x 108 m/s. Calculate its relativistic de Broglie wavelength. (Given m0 = 9.1 x 10-31 kg, $h = 6.63 x 10^{-34} J s$, $c = 3 x 10^{8} m/s)$
Calculate lambd$a = h / p = 6.63 \times 10^{-34} / (2.0475 \times 10^{-22}) = 3.24 \times 10^{-12} \text{ m}$.
Answer: 3.24 \times 10^{-12} \text{ m}
Assumptions and limitations
De Broglie wavelength formula assumes a uniform isotropic medium and free particle behavior.
Macroscopic objects have imperceptibly small de Broglie wavelengths because Planck's constant h is extremely small, making wave properties undetectable in classical mechanics.
Common misconceptions
Common error: Matter waves are electromagnetic waves because they propagate in vacuum.
Correct: Matter waves are probability waves (de Broglie waves), not electromagnetic waves, and do not involve oscillating electric and magnetic fields.
Electromagnetic waves are self-propagating transverse waves of electric and magnetic fields, whereas matter waves represent the probability amplitude of finding a particle.
Common error: The phase velocity of matter waves can transmit physical information faster than light.
Correct: The phase velocity vp exceeds the speed of light, but the group velocity vg (which represents particle velocity) is always less than c.
Information and energy are transported by the wave packet at the group velocity, not the individual phase ripples.
Quick revision
De Broglie wavelength is inversely proportional to momentum: lambd$a = h / p$.
For electrons, lambd$a = 12.27 / \sqrt{V} Angstroms$.
Davisson-Germer experiment proved the wave nature of electrons through crystal diffraction.
Group velocity of matter waves equals the particle velocity (vg = v).
Must remember
The de Broglie wavelength associated with a particle of mass $m$ and velocity $v$ is given by $\lambda = \frac{h}{mv}$.
For an electron accelerated through a potential $V$, the de Broglie wavelength is $\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}$.
The Davisson-Germer experiment confirmed the wave nature of electrons using nickel single crystal diffraction at $54\text{ V}$ and $50^{\circ}$.
The group velocity ($v_g$) of matter waves equals the classical particle velocity $v$, whereas the phase velocity ($v_p$) exceeds the speed of light ($c^2/v$).
Heisenberg's position-momentum uncertainty principle is expressed as $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.
Glossary
De Broglie WavelengthThe wavelength associated with a moving material particle, inversely proportional to its momentum.
Phase VelocityThe velocity with which a phase of a wave propagates; for matter waves, $v_p = \frac{c^2}{v} > c$.
Group VelocityThe velocity of the wave packet representing a moving particle, equal to the particle's classical velocity $v$.
Heisenberg Uncertainty PrincipleAn intrinsic fundamental limit to the simultaneous precision of conjugate dynamical variables.
Phase Velocity vs Group Velocity of Matter Waves
Parameter
Phase Velocity ($v_p$)
Group Velocity ($v_g$)
Definition
Velocity of individual phase ripples
Velocity of the overall wave packet
Relation to particle velocity ($v$)
$v_p = \frac{c^2}{v}$
$v_g = v$
Magnitude relative to speed of light ($c$)
Always greater than $c$ ($v_p > c$)
Always less than or equal to $c$ ($v_g \le c$)
Physical significance
Carries no energy or information
Carries the particle, energy, and information
Practice Quiz
Choose an option to reveal the explanation.
1. What is the de Broglie wavelength of an electron accelerated from rest through a potential difference of $100\text{ V}$?
Explanation: Using the formula for non-relativistic electrons, $\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA} = \frac{12.27}{\sqrt{100}} \text{ \AA} = \frac{12.27}{10} = 1.227 \text{ \AA}.
2. The Davisson-Germer experiment established the wave nature of electrons by observing diffraction from which crystal?
Explanation: Davisson and Germer used a nickel single crystal to demonstrate electron diffraction, observing a sharp intensity maximum at $54\text{ V}$ and $50^{\circ}$ scattering angle.
3. How does the de Broglie wavelength $\lambda$ of a particle vary with its kinetic energy $E$?
Explanation: Since momentum $p = \sqrt{2mE}$, the de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$, implying $\lambda \propto \frac{1}{\sqrt{E}}$.
4. In the context of matter waves, the group velocity ($v_g$) is related to the particle's classical velocity ($v$) by:
Explanation: The wave packet representing a moving material particle travels at the group velocity, which is identically equal to the classical particle velocity $v$.
5. What is the phase velocity ($v_p$) of a matter wave associated with a particle moving with velocity $v$ in free space?
Explanation: Phase velocity is given by $v_p = \frac{\omega}{k}$. For matter waves, this evaluates to $v_p = \frac{c^2}{v}$, which is greater than the speed of light $c$ since $v < c$.
6. Heisenberg's uncertainty principle relating position and momentum is mathematically expressed as:
Explanation: The standard minimum uncertainty relation for position and momentum is $\Delta x \cdot \Delta p \ge \frac{\hbar}{2}$ which is equivalent to $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.
7. If the momentum of a particle is doubled, its de Broglie wavelength becomes:
Explanation: De Broglie wavelength is inversely proportional to momentum ($\lambda = h/p$). If momentum $p$ is doubled, $\lambda$ is halved.
8. What characterizes matter waves as opposed to electromagnetic waves?
Explanation: Matter waves are probability waves representing the probability amplitude of finding a particle, unlike electromagnetic waves which are transverse oscillations of electric and magnetic fields.
9. At absolute temperature $T$, the root-mean-square thermal de Broglie wavelength of gas molecules of mass $m$ is proportional to:
Explanation: From the thermal wavelength formula $\lambda = \frac{h}{\sqrt{3mk_B T}}$, it is clear that $\lambda \propto \frac{1}{\sqrt{T}}$.
10. Which of the following particles, all moving with the same kinetic energy, has the longest de Broglie wavelength?
Explanation: Since $\lambda = \frac{h}{\sqrt{2mE}}$, for a given kinetic energy $E$, the particle with the smallest mass $m$ has the longest wavelength. The electron has the smallest mass among the choices.
11. The time-energy uncertainty relation is given by:
Explanation: The time-energy uncertainty relation derived from wave mechanics is $\Delta E \cdot \Delta t \ge \frac{h}{4\pi}$ (or $\ge \frac{\hbar}{2}$).
12. In the Davisson-Germer experiment, electrons were accelerated through a potential of:
Explanation: The sharp maximum in scattered electron intensity was observed at an accelerating potential of $54\text{ V}$.
13. Why can we not observe wave properties for a moving car on a highway?
Explanation: Due to Planck's constant $h$ being extremely small ($~10^{-34}\text{ J s}$), macroscopic objects with large masses have wavelengths so vanishingly small that wave interference or diffraction effects are completely unobservable.
14. Bragg's law for crystal diffraction is written as $2d \sin\theta = n\lambda$. Here, $\theta$ represents:
Explanation: In Bragg's law as applied to electron diffraction, $\theta$ is defined as the glancing angle made by the incident beam with the atomic planes of the crystal.
15. Which statement regarding phase velocity is correct?
Explanation: Although phase velocity $v_p = c^2/v > c$, this does not violate special relativity because phase ripples do not carry energy, mass, or information.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The de Broglie wavelength of an electron of kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $4E$, what will be the new de Broglie wavelength?
Explanation: Since $\lambda \propto \frac{1}{\sqrt{E}}$, when energy increases by a factor of 4, the wavelength decreases by a factor of $\sqrt{4} = 2$, becoming $\frac{\lambda}{2}$.
2. The ratio of de Broglie wavelengths of a proton and an $\alpha$ particle accelerated through the same potential difference $V$ is:
Explanation: De Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mqV}}$. For proton ($m_p, q_p = e$) and $\alpha$ particle ($m_\alpha = 4m_p, q_\alpha = 2e$): $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} = 2\sqrt{2} : 1$.
3. What is the scattering angle $\theta$ at which the maximum intensity of diffracted electrons was observed in the Davisson-Germer experiment?
Explanation: In the Davisson-Germer experiment, the sharp peak in electron scattering intensity was observed at an accelerating potential of $54\text{ V}$ and a scattering angle of $50^{\circ}$.
4. A particle of mass $m$ is moving with a velocity equal to $0.6\text{ }c$. The ratio of its phase velocity ($v_p$) to its group velocity ($v_g$) is:
5. If the uncertainty in the position of a particle is equal to its de Broglie wavelength, the minimum uncertainty in its momentum is approximately equal to:
Explanation: From Heisenberg's principle, $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. Given $\Delta x = \lambda = \frac{h}{p}$, we get $\left(\frac{h}{p}\right) \Delta p \ge \frac{h}{4\pi} \implies \Delta p \ge \frac{p}{4\pi}$. (Using exact minimal packet relation $\Delta x \Delta p \ge \hbar/2$, $\Delta p = \frac{\hbar}{2\lambda} = \frac{h}{4\pi \lambda} = \frac{p}{4\pi}$ or $\frac{p}{2\pi}$ depending on convention; standard minimum wave packet yields $\Delta x \Delta p = \hbar/2 \implies \Delta p = \frac{p}{4\pi}$ or $\frac{h}{2\lambda}$ scaled variants).
6. The de Broglie wavelength of a neutron at absolute temperature $T$ is proportional to:
Explanation: Thermal neutrons have wavelength given by $\lambda = \frac{h}{\sqrt{3mk_B T}}$, showing that $\lambda \propto T^{-1/2}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following relations correctly defines the de Broglie wavelength in terms of momentum $p$?
Explanation: De Broglie's fundamental hypothesis states that $\lambda = \frac{h}{p}$.
2. What is the phase velocity of a matter wave when the particle velocity approaches the speed of light ($v \to c$)?
Explanation: Since $v_p = \frac{c^2}{v}$, as $v \to c$, the phase velocity $v_p$ approaches $c$.
3. An electron and a proton have the same kinetic energy. Compare their de Broglie wavelengths.
Explanation: Since $\lambda = \frac{h}{\sqrt{2mE}}$ and $m_e < m_p$, the electron has a larger de Broglie wavelength than the proton.
4. 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 dissertation in 1924.
5. What is the product of phase velocity and group velocity for a matter wave in free space?
Explanation: Since $v_p = \frac{c^2}{v}$ and $v_g = v$, their product is $v_p \cdot v_g = \left(\frac{c^2}{v}\right)(v) = c^2$.
6. If the kinetic energy of a free particle is increased by a factor of 9, its de Broglie wavelength changes by a factor of:
Explanation: Since $\lambda \propto \frac{1}{\sqrt{E}}$, increasing energy by 9 times reduces the wavelength by a factor of $\sqrt{9} = 3$, i.e., $1/3$.
7. Which experiment provided direct confirmation of de Broglie's hypothesis of matter waves?
Explanation: The Davisson-Germer experiment provided the first experimental proof of the wave nature of electrons through crystal diffraction.
8. According to Heisenberg's uncertainty principle, if the position of a particle is known with absolute precision ($\Delta x = 0$), what is the uncertainty in its momentum?
Explanation: Since $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$, if $\Delta x = 0$, then $\Delta p$ must be infinite.
9. Matter waves are best described as:
Explanation: Matter waves are probability waves (associated with wave functions in quantum mechanics) that govern the probability of finding a particle in space.
10. What is the de Broglie wavelength of an $\alpha$ particle accelerated through a potential difference of $V$?
Explanation: For an $\alpha$ particle ($m = 4m_p, q = 2e$), $\lambda = \frac{h}{\sqrt{2mqV}} = \frac{0.101}{\sqrt{V}} \text{ \AA}$ (using standard numerical calculation for $\alpha$ particles).