A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is pr

NCERT Exemplar · Q1

H

(a)

H

(b)

H^{1/2}

(c)

H^{0}

(d)

H^{-1/2}

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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).

Important

De Broglie's hypothesis is not just a clever idea — it's the foundation of quantum mechanics. Every particle has a wavelength, and that wavelength determines how it moves, where it can be found, and even why electrons in atoms occupy only certain discrete energy levels (standing waves around the nucleus).

A Quick Way to Remember

For an exam, you'll often need to compute the de Broglie wavelength of an electron accelerated through a potential difference $V$ volts. The kinetic energy gained is $eV$ , so:

\frac{1}{2}mv^2 = eV \quad \Rightarrow \quad v = \sqrt{\frac{2eV}{m}}

Substituting into $\lambda = h/(mv)$ gives:

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

Plug in the numbers ( $h$ , $m_e$ , $e$ ) and you get a handy formula:

Tip

For an electron accelerated through $V$ volts:

$\lambda (\text{in } \text{Å}) = \frac{12.27}{\sqrt{V}}$

So a 100 V electron has $\lambda \approx 1.23 \ \text{Å}$ — right in the X-ray range.

The Bottom Line

De Broglie wavelength is the bridge between the particle and wave pictures of matter. It tells you that momentum and wavelength are two sides of the same coin. For large objects, the wavelength is negligible — Newtonian physics works fine. For tiny particles, the wavelength dominates — and you must use quantum mechanics.

When you see $\lambda = h/p$ , remember: that's nature saying that everything — from electrons to planets — has a wave nature. It's just that for most things, the wave is too small to notice.

Searches like "de Broglie wavelength formula and examples" and "dual nature of matter class 12 physics" are very common, since this concept is central to the Dual Nature of Radiation and Matter chapter of the NCERT/CBSE Class 12 Physics curriculum. The handy $\lambda = 12.27/\sqrt{V}$ shortcut for accelerated electrons is a frequent JEE Main and NEET numerical question.

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:

Davisson-Germer experiment (1927): Electrons scattered off a nickel crystal produced diffraction patterns — exactly like X-rays (waves!)
The measured wavelength matched $\lambda = h/p$ perfectly

This was Nobel Prize material — de Broglie won in 1929.

Key Takeaways for Exams

Concept	Formula	When to Use
De Broglie wavelength	$\lambda = \frac{h}{p}$	Always — fundamental definition
Non-relativistic	$\lambda = \frac{h}{mv}$	For $v \ll c$ (most exam problems)
Relativistic	$\lambda = \frac{h}{\gamma mv}$	For $v \approx c$ (rare in school exams)
For an electron accelerated through $V$ volts	$\lambda = \frac{h}{\sqrt{2meV}}$	Derive from $p = \sqrt{2mE_k}$

The Deeper "Why" — One Sentence

De Broglie wavelength exists because nature is symmetric: just as light has both wave and particle aspects, so must matter — and the bridge between them is Planck's constant $h$ .

The formula $\lambda = h/p$ is not derived from deeper principles — it is the fundamental postulate that connects the particle's momentum to its wave's wavelength. Its validity comes from experiment, not from pure mathematics.

A particle dropped from height $H$ reaches speed $v=\sqrt{2gH}$ once it has fallen through the full height $H$ ; since $\lambda = h/p$ and $p=mv$ , the de Broglie wavelength is proportional to $H^{-1/2}$ — option (D).

The de Broglie wavelength of any moving particle is $\lambda = h/p$ , where $h$ is Planck's constant and $p$ is the particle's linear momentum. So the problem reduces to finding how the particle's momentum depends on the drop height $H$ .

Set up the free fall. The particle starts at rest and falls under gravity through the full height $H$ . Using $v^2 = u^2 + 2gs$ with $u=0$ and $s=H$ (the distance covered once it has fallen the whole way):

v = \sqrt{2gH}.

Find the momentum.

p = mv = m\sqrt{2gH}.

Find the de Broglie wavelength.

\lambda = \frac{h}{p} = \frac{h}{m\sqrt{2gH}}.

Holding the mass $m$ and $g$ fixed, this gives

\lambda \propto \frac{1}{\sqrt{H}} = H^{-1/2}.

Watch out

Don't confuse Planck's constant $h$ with the particle's height — here always called $H$ . Keeping them as distinct symbols throughout avoids the derivation collapsing into an ambiguous $h$ -vs- $h$ mix-up.

Check the other options.
- (A) $H$ : would mean $\lambda$ grows with drop height — wrong, since a bigger drop gives a bigger speed and hence a shorter wavelength.
- (B) $H^{1/2}$ : the opposite dependence to the correct one.
- (C) $H^{0}$ : constant — only true if speed didn't depend on $H$ at all, which is false since $v=\sqrt{2gH}$ .
- (D) $H^{-1/2}$ : matches the derivation above.

✓Final answer

\lambda \propto H^{-1/2}

Method: Finding How de Broglie Wavelength Depends on a Changing Parameter

This method applies whenever a particle's speed changes due to some physical process (falling under gravity, acceleration by a field, etc.) and you're asked how its de Broglie wavelength depends on a variable describing that process.

Steps

Step 1: Identify the physical process that fixes the momentum

Before touching $\lambda = h/p$ , work out what mechanics principle governs the particle's speed at the point of interest — energy conservation (e.g. free fall, $\frac{1}{2}mv^2 = mg\times(\text{height fallen})$ ), Newton's second law with a constant force, or a given kinetic energy. This is the step most students skip, and it's the one that actually determines the answer.

Step 2: Express momentum as a function of the given variable

Use the mechanics relation from Step 1 to write speed (and hence momentum $p = mv$ ) purely in terms of the variable the question asks about — here, a height. Keep the constants ( $m$ , $g$ ) symbolic; you only need their combination, not their values.

p = mv = m\sqrt{2g\,(\text{height fallen})}

Step 3: Substitute into $\lambda = h/p$ and reduce to a proportionality

\lambda = \frac{h}{p} \ \propto\ \frac{1}{\sqrt{\text{height fallen}}}

Since only the proportionality is asked (not a numeric value), drop every constant ( $h$ , $m$ , $g$ ) and keep only how $\lambda$ scales with the variable.

Step 4: Read the question's variable carefully before matching an option

Exam questions like this often use two different heights in the same sentence (the drop height and the instantaneous height) — check exactly which one the option is asking about, since $\lambda \propto (\text{drop height})^{-1/2}$ is the same power-law shape you'd get with any variable that momentum scales as its square root, so a careless reading can pick the wrong option letter even with all the algebra done right.

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

KEAM 2026Set eng-2026-04174 marksMCQ
$\lambda_e$
›Reveal solutionSolution

For equal kinetic energy the de Broglie wavelength varies as $1/\sqrt{m}$ ; the lighter electron has the longer wavelength.

The de Broglie wavelength in terms of kinetic energy $K$ :
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}.$
With the same $K$ for both particles, $\lambda \propto \dfrac{1}{\sqrt{m}}$ . Since $m_e \ll m_p$ (proton is $\approx 1836$ times heavier):
$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1836} > 1 \quad\Rightarrow\quad \lambda_e > \lambda_p.$
(Note option C would need $\lambda_e = \sqrt{1836}\,\lambda_p$ , the reverse of what it states, so it is wrong.)

✓Final answer
$The correct option is (A) .$
KEAM 2026Set pha-2026-0418F4 marksMCQ
$Q. The wavelength of a fast moving particle (λ) is related to its momentum (p) and Planck's constant$
(h) as (A) $\lambda = hp$ (B) $\lambda = h^{2} / p$ (C) $\lambda = h / p$ (D) $\lambda = p^{2} / h$ (E) $\lambda = h / p^{2}$
›Reveal solutionSolution

The de Broglie wavelength is $\lambda=h/p$ .

Reasoning

de Broglie proposed that a moving particle has an associated wavelength
$\lambda=\frac{h}{p}=\frac{h}{mv},$
inversely proportional to momentum. This matches option (C).

✓Final answer
$The correct option is (C) .$
KEAM 2026Set pha-2026-0418F4 marksMCQ
$Q. If the de Broglie wavelengths of proton p and alpha particle α are same, then (A) both have same momentum (B) both have same energy (C) momentum of p is twice that of α (D) momentum of α is twice that of p (E) energy of p is twice that of α$
›Reveal solutionSolution

$\lambda = h/p$ ; same de Broglie wavelength $\Rightarrow$ same momentum.

The de Broglie wavelength is $\lambda = \dfrac{h}{p}$ , so momentum $p = h/\lambda$ . If the proton and alpha particle have the same $\lambda$ , they must have the same momentum $p$ . (Their energies differ, since $E = p^2/2m$ and the alpha particle is about four times heavier, but momentum is equal.)

✓Final answer
$The correct option is (A) .$
KEAM 2025Set eng-2025-04254 marksMCQ
$m$
›Reveal solutionSolution

The de Broglie wavelength in terms of kinetic energy is $\lambda=h/\sqrt{2mK}$ ; substituting $K=2E$ gives $h/(2\sqrt{mE})$ .

Momentum related to kinetic energy: $p=\sqrt{2mK}$ . Here $K=2E$ , so
$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}=\dfrac{h}{\sqrt{2m(2E)}}=\dfrac{h}{\sqrt{4mE}}=\dfrac{h}{2\sqrt{mE}}.$
✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04264 marksMCQ
$\lambda$
›Reveal solutionSolution

The de Broglie wavelength of an accelerated electron varies as $1/\sqrt{V}$ ; reducing $V$ from 480 V to 120 V (factor 4) doubles the wavelength.

An electron accelerated through potential $V$ has de Broglie wavelength
$\lambda = \frac{h}{\sqrt{2meV}} \propto \frac{1}{\sqrt{V}}$
Taking the ratio for the two potentials:
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2$
Since $\lambda_1 = \lambda$ , we get $\lambda_2 = 2\lambda$ .

✓Final answer
$The correct option is (C) .$
KEAM 2024Set eng-2024-06084 marksMCQ
$\lambda_p$
›Reveal solutionSolution

A photon's energy varies as $1/\lambda_0$ , while a proton's kinetic energy varies as $1/\lambda_p^{2}$ . Equal energies force $\lambda_0\propto\lambda_p^{2}$ .

For the photon, energy and wavelength are related by
$E=\frac{hc}{\lambda_0}\ \Rightarrow\ E\propto\frac{1}{\lambda_0}.$
For the (non-relativistic) proton, the de Broglie wavelength is $\lambda_p=\frac{h}{p}$ , and its kinetic energy is
$E=\frac{p^2}{2m}=\frac{h^2}{2m\lambda_p^{2}}\ \Rightarrow\ E\propto\frac{1}{\lambda_p^{2}}.$
Setting the two energies equal, $\frac{1}{\lambda_0}\propto\frac{1}{\lambda_p^{2}}$ , hence
$\lambda_0\propto\lambda_p^{2}.$
✓Final answer
$The correct option is (D) .$
KEAM 2024Set pha-2024-06104 marksMCQ
$Q. If the speed of a moving particle is decreased by 1%, the de Broglie wavelength of the wave associated with it (A) decreases by 1% (B) increases by 1% (C) decreases by 2% (D) increases by 2% (E) decreases by 5%$
›Reveal solutionSolution

de Broglie wavelength is inversely proportional to speed.

The de Broglie wavelength is
$\lambda = \frac{h}{mv}\;\Rightarrow\; \lambda \propto \frac{1}{v}.$
For small changes, $\frac{\Delta\lambda}{\lambda} = -\frac{\Delta v}{v}$ . A $1\%$ decrease in $v$ ( $\Delta v/v = -1\%$ ) gives $\Delta\lambda/\lambda = +1\%$ , an increase of about 1%.

✓Final answer
$The correct option is (B) . Inverse proportionality means a 1% speed decrease increases the wavelength by ~1%.$
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$\alpha$
›Reveal solutionSolution
The de-Broglie wavelength ratio $\lambda_\alpha:\lambda_p$ is $2:1$ .

Concept and Intuition

The de-Broglie wavelength is inversely proportional to momentum, $\lambda=h/p$ . Halving the momentum doubles the wavelength.

Step-by-Step Solution

$\dfrac{\lambda_\alpha}{\lambda_p}=\dfrac{h/p_\alpha}{h/p_p}=\dfrac{p_p}{p_\alpha}$ .

Given $p_\alpha=\dfrac{p_p}{2}$ .

$\dfrac{\lambda_\alpha}{\lambda_p}=\dfrac{p_p}{p_p/2}=2$ , so the ratio is $2:1$ .

Common Mistakes

Treating wavelength as directly proportional to momentum.

✓Final answer
$2:1$

ANSWER: E

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