JEE Mains Physics · Dual Nature of Radiation and Matter

de Broglie Wavelength of a Particle

Every moving particle has a wavelength λ = h/p = h/√(2mK); for a charge q accelerated from rest through V it is h/√(2mqV), so the wavelength falls as the speed, the energy or the voltage rises.

All Dual Nature of Radiation and Matter notes JEE Mains Physics strategy

Why this matters

Twenty-four PYQs, twenty-two of them multiple choice, and five from 2026. Fifteen change one thing and ask how the wavelength follows: five an accelerating voltage, three a kinetic energy, two a speed, three a gas temperature, one a photoelectron's energy and one a Bohr orbit. Four put an electron in an electric or a magnetic field. Five ask about the evidence for matter waves: the Davisson–Germer experiment, the electron microscope and the uncertainty principle.

Concept 1 of 3: de Broglie wavelength and how it scales

de Broglie's idea is that a particle has a wavelength set by its momentum alone: λ = h/p. Everything else is a way of writing the momentum. From the kinetic energy, p = √(2mK). From an accelerating voltage, K = qV. From a temperature, K = 3kT/2. So the wavelength falls as 1/v, but only as 1/√K or 1/√V.

Definition

$\lambda = \dfrac{h}{p} = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mqV}}$ .
Electron accelerated from rest through V volts: $\lambda = \dfrac{1.227}{\sqrt{V}}$ nm (or $12.27/\sqrt{V}$ Å).
So $\lambda \propto 1/v$ , $\lambda \propto 1/\sqrt{K}$ and $\lambda \propto 1/\sqrt{V}$ . For one particle, $\lambda^{2}K = h^{2}/2m$ stays fixed.
A gas molecule at temperature T, taking $K = \tfrac{3}{2}kT$ : $\lambda = \dfrac{h}{\sqrt{3mkT}}$ , so $\lambda \propto 1/\sqrt{T}$ .
A photoelectron: $\lambda_e = \dfrac{h}{\sqrt{2mK_{\max}}}$ , with $K_{\max}$ from Einstein's equation.
An electron in the nth Bohr orbit of radius r fits n whole waves round it: $n\lambda = 2\pi r$ .

de Broglie wavelength

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}, \qquad \lambda_{e} = \frac{1.227}{\sqrt{V}}\ \text{nm}

Worked example

An electron is accelerated from rest through 150 V. Find its de Broglie wavelength. Through what voltage must it be accelerated to make its wavelength one third of this?

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The same idea in a real exam question:

JEE Mains · 2021 · Paper 19 · Q2Moderate

Example 1 · Dual Nature of Radiation and Matter · de Broglie Wavelength of a Particle

The de-Broglie wavelength of a particle having kinetic energy

E

\lambda

. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to

75\%

of the initial value?

λ goes as 1/√K, not 1/K

Momentum is √(2mK). Four times the kinetic energy halves the wavelength; it does not quarter it. The same holds for the accelerating voltage.

Use the particle's own charge

A charge q accelerated through V gains qV. An alpha particle has charge 2e, so through V it gains 2eV, not eV.

Extra energy is the change, not the new total

When a question asks how much energy must be ADDED, find the new kinetic energy and subtract the old one. The new total is a distractor.

Drill 14 more on de broglie wavelength and how it scales

Concept 2 of 3: de Broglie wavelength of an electron in an electric or magnetic field

The wavelength follows the speed. So find how the field changes the speed, then use λ = h/mv. An electric field along the motion speeds the particle up or slows it down. One at right angles adds a sideways velocity, which also raises the speed. A magnetic force is always at right angles to the motion, so it turns the particle but never changes its speed.

Definition

Find $\vec v(t)$ , then $\lambda(t) = \dfrac{h}{m\,|\vec v(t)|}$ .
An electron has charge −e, so the force on it is $\vec F = -e\vec E$ , opposite to the field.
Field along the motion: $v = v_0 \pm \dfrac{eE}{m}t$ , with the sign from the direction of the force.
Field at right angles: $|\vec v| = \sqrt{v_0^{2} + \left(\dfrac{eEt}{m}\right)^{2}}$ , so λ falls.
A magnetic force does no work. The speed, and so λ, stays the same.

Wavelength in a field

\lambda(t) = \frac{h}{m\,|\vec v(t)|}, \qquad \vec F = -e\vec E \ \text{(electron)}

Worked example

An electron of mass m starts with velocity

v_0\hat i

and de Broglie wavelength

\lambda_0

. A uniform field

\vec E = E_0\hat i

(

E_0 > 0

) acts on it. Find its wavelength at time t, and the time at which the wavelength becomes

2\lambda_0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q17Moderate

Example 2 · Dual Nature of Radiation and Matter · de Broglie Wavelength of a Particle

An electron (mass

m

) with an initial velocity

\overrightarrow{v}=v_{0}\widehat{i}\left( v_{0}> 0 \right)

is moving in an electric field

\overrightarrow{E}= -E_{0}\widehat{i}\left( E_{0}> 0 \right)

where

E_{0}

is constant. If at

t = 0

de Broglie wavelength is

\lambda_{0}=\frac{h}{mv_{0}}

, then its de Broglie wavelength after time

t

is given by

The force on an electron is opposite to the field

For an electron, F = −eE. A field pointing against the motion pushes the electron forward and shortens its wavelength; a field along the motion slows it and lengthens the wavelength.

A sideways electric field still changes λ

A field at right angles to the motion adds a sideways velocity. The speed grows as √(v₀² + (eEt/m)²), so the wavelength falls, even though the original velocity component is unchanged.

A magnetic field never changes λ

The magnetic force is always at right angles to the velocity, so it does no work and the speed stays the same. The wavelength is unchanged at every instant.

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Concept 3 of 3: Evidence for matter waves

Light behaves as a wave in interference and as particles in the photoelectric effect. de Broglie said matter does the same in reverse. Electrons sent through a crystal make a diffraction pattern, just as X-rays do, and the wavelength measured agrees with h/p. The short wavelength of fast electrons is what makes an electron microscope so sharp.

Definition

Every moving particle has a wavelength λ = h/p. Light and matter both show wave and particle behaviour.
Interference and diffraction are the signature of a wave; the photoelectric effect is the signature of particles.
Matter waves are not electromagnetic waves; a neutral particle has one too.
The finest detail a microscope can show scales with the wavelength it uses, so resolving power $\propto 1/\lambda$ .
Heisenberg's uncertainty principle: $\Delta x\,\Delta p \ge \dfrac{h}{4\pi}$ .

Observation or device	What it shows	Key relation
Davisson–Germer experiment	Electrons scattered from a nickel crystal give a diffraction peak, so electrons behave as waves	At 54 V the peak is at 50°; the measured λ ≈ 0.165 nm matches h/p
Electron diffraction and interference	A beam of electrons spreads and makes fringes, like light	Fringe spacing grows with λ = h/p
Electron microscope	Resolves far finer detail than an optical microscope	Electron λ is a fraction of a nanometre, against 400–700 nm for light
Heavier particle at the same speed	Shorter wavelength, finer detail	λ = h/mv, so at equal speed λ ∝ 1/m
Photoelectric effect	Light arrives as particles, photons	E = hν per photon
Heisenberg uncertainty principle	Position and momentum cannot both be known exactly	Δx Δp ≥ h/4π
Everyday objects	No visible wave effects	For a large mass, h/mv is far smaller than any gap or slit

Wave behaviour is shown by diffraction and interference; particle behaviour by one-at-a-time energy exchange.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 25 · Q4Moderate

Example 3 · Dual Nature of Radiation and Matter · de Broglie Wavelength of a Particle

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: An electron microscope can achieve better resolving power than an optical microscope. Reason R: The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light. In the light of the above statements, choose the correct answer from the options given below:

Matter waves are not electromagnetic

An electron's wave is not light and does not need a charge. A neutron or a whole atom has a de Broglie wavelength too.

Diffraction means wave, photoelectric means particle

Electron diffraction is evidence for the wave nature of matter. The photoelectric effect is evidence for the particle nature of light. Swapping the two is a common wrong statement.

Drill 4 more on evidence for matter waves

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

de Broglie wavelength and how it scales
de Broglie wavelength
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}, \qquad \lambda_{e} = \frac{1.227}{\sqrt{V}}\ \text{nm}$
de Broglie wavelength of an electron in an electric or magnetic field
Wavelength in a field
$\lambda(t) = \frac{h}{m\,|\vec v(t)|}, \qquad \vec F = -e\vec E \ \text{(electron)}$

Reference tables (1)

Evidence for matter waves7 rows

Observation or device	What it shows	Key relation
Davisson–Germer experiment	Electrons scattered from a nickel crystal give a diffraction peak, so electrons behave as waves	At 54 V the peak is at 50°; the measured λ ≈ 0.165 nm matches h/p
Electron diffraction and interference	A beam of electrons spreads and makes fringes, like light	Fringe spacing grows with λ = h/p
Electron microscope	Resolves far finer detail than an optical microscope	Electron λ is a fraction of a nanometre, against 400–700 nm for light
Heavier particle at the same speed	Shorter wavelength, finer detail	λ = h/mv, so at equal speed λ ∝ 1/m
Photoelectric effect	Light arrives as particles, photons	E = hν per photon
Heisenberg uncertainty principle	Position and momentum cannot both be known exactly	Δx Δp ≥ h/4π
Everyday objects	No visible wave effects	For a large mass, h/mv is far smaller than any gap or slit

Wave behaviour is shown by diffraction and interference; particle behaviour by one-at-a-time energy exchange.

Watch out for (8)

λ goes as 1/√K, not 1/K
→ de Broglie wavelength and how it scales
Use the particle's own charge
→ de Broglie wavelength and how it scales
Extra energy is the change, not the new total
→ de Broglie wavelength and how it scales
The force on an electron is opposite to the field
→ de Broglie wavelength of an electron in an electric or magnetic field
A sideways electric field still changes λ
→ de Broglie wavelength of an electron in an electric or magnetic field
A magnetic field never changes λ
→ de Broglie wavelength of an electron in an electric or magnetic field
Matter waves are not electromagnetic
→ Evidence for matter waves
Diffraction means wave, photoelectric means particle
→ Evidence for matter waves

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