An electron (mass $m$) with an initial velocity $\vec{v} = v_0\,\hat{i}$ is in an electric field $\vec{E} = E_0\,\hat{j}$. If $\lambda_0 = h/m v_0$, its de Broglie wavelength at time $t$ is given by (a) $\lambda_0$ (b) $\lambda_0 \sqrt{1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}}$ (c) $\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$ (d) $\dfrac{\lambda_0}{\left(1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}\right)}$

An electron (mass m) with an initial velocity vecv = v₀hati is in an electric field vecE = E₀hatj. If λ₀ = h/m

NCERT Exemplar · Q7

m

(a)

\lambda_0

(b)

\lambda_0 \sqrt{1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}}

(c)

\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}}}

(d)

\dfrac{\lambda_0}{\left(1 + \dfrac{e^2 E_0^2 t^2}{m^2 v_0^2}\right)}

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Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

Imagine you're holding a cricket ball. You know exactly where it is, and if you throw it, you can predict its path. That's a particle — localised, definite, following Newton's laws. Now think of light. You can't "hold" a beam of light; it spreads out, bends around corners, creates interference patterns. That's a wave — spread out, not localised.

For centuries, physics kept these two worlds separate. Particles were particles. Waves were waves. Never the twain shall meet.

Then came a young French physicist, Louis de Broglie, in 1924. He asked a question that seemed almost absurd: If light — which we thought was a wave — can behave like a particle (the photoelectric effect), then why can't a particle — say, an electron — behave like a wave?

That question turned physics upside down.

The Core Idea

De Broglie proposed that every moving particle has a wave associated with it. The wavelength of that wave depends on the particle's momentum. The faster or heavier the particle, the shorter the wavelength.

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

This is not a mathematical trick. It's a physical reality. An electron moving through a crystal actually behaves like a wave of this wavelength — it can diffract, interfere, and form patterns just like light does.

Why You Don't See It in Daily Life

Here's the crucial point: the de Broglie wavelength is incredibly tiny for everyday objects.

Take a cricket ball of mass 0.16 kg moving at 30 m/s. Its de Broglie wavelength is:

\lambda = \frac{6.626 \times 10^{-34}}{0.16 \times 30} \approx 1.38 \times 10^{-34} \ \text{m}

That's about a hundred trillion trillion times smaller than the nucleus of an atom. No experiment can detect such a wave — it's effectively zero for all practical purposes.

Now take an electron (mass $9.1 \times 10^{-31}$ kg) accelerated through 100 volts. Its speed is about $5.9 \times 10^6$ m/s. Its de Broglie wavelength:

\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.9 \times 10^6} \approx 1.23 \times 10^{-10} \ \text{m}

That's about 0.12 nanometres — comparable to the spacing between atoms in a crystal. This is measurable. And indeed, in 1927, Davisson and Germer fired electrons at a nickel crystal and observed diffraction — the unmistakable signature of a wave.

Note

The de Broglie wavelength is only observable when it is comparable to the size of objects the particle interacts with. For macroscopic objects, it's far too small to matter. For subatomic particles, it's the key to understanding their behaviour.

What This Means Physically

The wave is not a physical wave in space like a water wave. It's a probability wave — its amplitude at any point tells you the probability of finding the particle there. Where the wave amplitude is large, you're likely to find the particle; where it's zero, you won't.

This wave-particle duality is not a compromise. It's the actual nature of reality. An electron is neither a pure particle nor a pure wave — it's something that shows particle-like behaviour in some experiments (like hitting a screen at a point) and wave-like behaviour in others (like passing through two slits and interfering with itself). …

Why this formula?

De Broglie Wavelength: Why the Formula Holds

Let's build this from the ground up — understanding why matter has a wavelength, not just memorizing $\lambda = \frac{h}{p}$ .

The Core Insight: Nature's Symmetry

Before de Broglie, physics had two separate worlds:

Light — showed wave behaviour (diffraction, interference) but also particle behaviour (photoelectric effect)
Matter — showed particle behaviour (momentum, collisions) but no wave behaviour yet

De Broglie asked a daring question in his 1924 PhD thesis:

If light (a wave) can behave like a particle, why can't a particle (like an electron) behave like a wave?

Nature should be symmetric — what applies to one should apply to the other.

Step 1: Start with Light (What We Already Knew)

For a photon, Einstein had given us two key relations:

Energy: $E = hf$ (Planck's relation)
Momentum: $p = \frac{h}{\lambda}$ (from $E = pc$ for light, combined with $c = f\lambda$ )

So for light:

\lambda = \frac{h}{p}

This was experimentally verified for photons.

Step 2: De Broglie's Bold Hypothesis

De Broglie said: This relation is not special to light. It is universal.

For any particle with momentum $p$ :

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

Where:

$\lambda$ = de Broglie wavelength
$h$ = Planck's constant ( $6.626 \times 10^{-34} \text{ J·s}$ )
$p$ = momentum of the particle

Step 3: Why Momentum and Not Velocity?

This is crucial. The formula uses momentum ( $p = mv$ ), not just velocity.

For a non-relativistic particle (slow compared to light):

\lambda = \frac{h}{mv}

For a relativistic particle (like an electron at high speed):

p = \gamma mv \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}

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

Why momentum? Because momentum is the more fundamental quantity — it's conserved, it's frame-independent in a deeper sense, and it connects directly to the wave's phase.

Step 4: The Deeper Reasoning — Wave-Particle Duality

De Broglie didn't just guess. He reasoned:

Every moving particle has an associated wave — called the "matter wave" or "pilot wave"
The frequency of this wave comes from energy: $f = \frac{E}{h}$
The wavelength comes from momentum: $\lambda = \frac{h}{p}$

These two relations are linked by the phase velocity of the wave:

v_{\text{phase}} = f\lambda = \frac{E}{h} \cdot \frac{h}{p} = \frac{E}{p}

For a free particle with kinetic energy $E = \frac{p^2}{2m}$ :

v_{\text{phase}} = \frac{p}{2m} = \frac{v}{2}

This is half the particle's speed — a strange but mathematically consistent result.

Step 5: Experimental Confirmation (Why We Believe It)

De Broglie's idea was confirmed when electrons showed wave behaviour: …

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The electron's charge is $-e$ , so the field $\vec{E}=E_0\hat{j}$ pushes it in the $-\hat{j}$ direction, building up a growing $y$ -component of velocity. Its speed - and hence its de Broglie wavelength - changes with time as $\lambda(t)=\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}$ , matching option (C).

Setting up the motion

The electron has charge $-e$ (with $e>0$ the elementary charge) and starts with velocity $\vec{v}(0)=v_0\hat{i}$ . In the field $\vec{E}=E_0\hat{j}$ , the force on it is

\vec{F} = (-e)\vec{E} = -eE_0\,\hat{j},

so the acceleration is $\vec{a} = -\dfrac{eE_0}{m}\hat{j}$ - the electron is pushed in the $-\hat{j}$ direction (opposite to $\vec{E}$ , since the charge is negative).

Velocity at time $t$

No force acts along $\hat{i}$ , so $v_x(t)=v_0$ stays constant. Along $\hat{j}$ :

v_y(t) = -\frac{eE_0}{m}\,t.

Speed (magnitude of velocity)

v(t) = \sqrt{v_x^2+v_y^2} = \sqrt{v_0^2 + \left(\frac{eE_0t}{m}\right)^2}.

(Note that the sign of $v_y$ doesn't matter here - only $v_y^2$ enters the speed, so whether the force pushes the electron in $+\hat{j}$ or $-\hat{j}$ , the final speed formula is identical. What matters for correctness is simply using the electron's true charge, $-e$ , in setting up the force in the first place.)

De Broglie wavelength …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL1 markMCQ
$Q. If λp and λα be the wavelength of de Broglie waves for a proton and an alpha particle then which of the following is correct? (Choose the correct option)$
(i) λp = λα
(ii) λp > λα
(iii) λp < λα
›Reveal solutionSolution

When accelerated through the same potential difference, the de Broglie wavelength of a proton is larger than that of an alpha particle, i.e. λp > λα.

The de Broglie wavelength of a particle of charge q and mass m, accelerated from rest through potential difference V, is λ = h/p = h/√(2mqV) (since qV = ½mv² = p²/2m).

For a proton: mass mp, charge e: λp = h/√(2mp·e·V).

For an alpha particle: mass ≈ 4mp, charge 2e: λα = h/√(2·4mp·2e·V) = h/√(16mp·e·V).

Taking the ratio: λp/λα = √(16mp·e·V / 2mp·e·V) = √8 ≈ 2.83.

…

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AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL1 mark
$Q. de Broglie in 1924 reasoned that nature was symmetrical and that the two basic physical entities ___ and ___ must have symmetrical character. (Fill in the blanks)$
›Reveal solutionSolution

De Broglie argued that since radiation (energy) shows dual particle-wave behaviour, matter should too, by symmetry of nature.

By 1924, it was firmly established that radiation (light) has a dual character — behaving as a wave in phenomena like interference and diffraction, and as a particle (photon) in phenomena like the photoelectric effect and Compton effect.

Louis de Broglie extended this idea by a bold hypothesis: nature loves symmetry, and if radiation (energy) can behave as both wave and particle, then matter (which was until then thought of only as particles) should also possess an associated wave character under suitable conditions. He proposed that the two basic physical entities of nature — matter and energy/radiation — must ha …

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←Previous questionQ6 — An electron (mass m) with an initial velocity vecv = v₀hati (v₀ > 0) i…→Next questionQ8 — A proton and an α-particle are accelerated, using the same potential d…

Q.An electron (mass $m$ ) with an initial velocity $\vec{v} = v_0\,\hat{i}$ is in an electric field $\vec{E} = E_0\,\hat{j}$ . If $\lambda_0 = h/m v_0$ , its de Broglie wavelength at time $t$ is given by

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Core Idea

Why You Don't See It in Daily Life

What This Means Physically

Why this formula?

De Broglie Wavelength: Why the Formula Holds

The Core Insight: Nature's Symmetry

Step 1: Start with Light (What We Already Knew)

Step 2: De Broglie's Bold Hypothesis

Step 3: Why Momentum and Not Velocity?

Step 4: The Deeper Reasoning — Wave-Particle Duality

Step 5: Experimental Confirmation (Why We Believe It)

Setting up the motion

Velocity at time $t$

Speed (magnitude of velocity)

De Broglie wavelength …

Method: Finding $\lambda(t)$ When a Force Acts Perpendicular to the Initial Velocity

Steps

Q.An electron (mass mmm) with an initial velocity v⃗=v0 i^\vec{v} = v_0\,\hat{i}v=v0​i^ is in an electric field E⃗=E0 j^\vec{E} = E_0\,\hat{j}E=E0​j^​. If λ0=h/mv0\lambda_0 = h/m v_0λ0​=h/mv0​, its de Broglie wavelength at time ttt is given by

Setting up the motion

Velocity at time ttt

Speed (magnitude of velocity)

De Broglie wavelength …

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Q.An electron (mass $m$ ) with an initial velocity $\vec{v} = v_0\,\hat{i}$ is in an electric field $\vec{E} = E_0\,\hat{j}$ . If $\lambda_0 = h/m v_0$ , its de Broglie wavelength at time $t$ is given by

Velocity at time $t$

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