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}

Gujarat GsebMCQ· 1mImportance★★★★★

49% · 41/83 Questions

✓ Free question

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.

GUJCET 2026Set x1 markMCQ
$\lambda$
›Reveal solutionSolution

For a photon the de Broglie wavelength equals the EM wavelength: $\lambda'=\lambda$ .

A photon of electromagnetic radiation of wavelength $\lambda$ carries momentum
$p=\frac{h}{\lambda}$
Its de Broglie wavelength is
$\lambda'=\frac{h}{p}=\frac{h}{h/\lambda}=\lambda$
So the de Broglie wavelength of the photon is exactly equal to the wavelength of the radiation.

✓Final answer
$\lambda'=\lambda$

ANSWER: (B)
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. The de Broglie wavelength of proton and alpha-particle is same. The ratio of their velocities is ___.$
(a) 4 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
›Reveal solutionSolution

Equal de Broglie wavelengths mean equal momenta (lambda = h/p), so the lighter particle must move faster in inverse proportion to the mass ratio.

lambda = h/(m v), so equal lambda means m_p v_p = m_alpha v_alpha.

Mass of alpha particle m_alpha = 4 m_p (2 protons + 2 neutrons, approximately 4 times proton mass).

m_p v_p = 4 m_p v_alpha => v_p/v_alpha = 4.

So v_p : v_alpha = 4 : 1.

✓Final answer
$(a) 4 : 1.$
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. A proton, a neutron, an electron and an alpha-particle have same energy. Then their de-Broglie wavelengths compare as$
(a) lambda_e = lambda_p = lambda_n = lambda_a
(b) lambda_e < lambda_p = lambda_n > lambda_a
(c) lambda_a < lambda_p = lambda_n < lambda_e
(d) lambda_p = lambda_n > lambda_e > lambda_a
›Reveal solutionSolution

At the same kinetic energy E, de Broglie wavelength lambda = h/sqrt(2 m E) is inversely proportional to sqrt(mass), so the heaviest particle has the smallest wavelength.

lambda = h / sqrt(2 m E). For fixed E, lambda ~ 1/sqrt(m).

Masses: m_e (electron) is by far the smallest; m_p (proton) approx equals m_n (neutron); m_alpha (alpha particle) approx 4 m_p, the largest.

So, ordering by increasing mass: electron < proton = neutron < alpha.

Hence ordering by increasing wavelength (inverse of mass order): alpha (smallest lambda) < proton = neutron < electron (largest lambda).

This matches lambda_alpha < lambda_p = lambda_n < lambda_e.

✓Final answer
$(c) lambda_alpha < lambda_p = lambda_n < lambda_e.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If de-Broglie wavelength of a dust particle of mass 1.0 x 10^-9 kg is 3 x 10^-25 m then the speed of the particle is ___. (h = 6.625 x 10^-34 Js)$
(a) 1.1 ms^-1
(b) 1.0 kms^-1
(c) 1.2 kms^-1
(d) 2.2 ms^-1
›Reveal solutionSolution

de Broglie wavelength: λ = h/(mv), so v = h/(mλ).

m = 1.0 × 10⁻⁹ kg, λ = 3 × 10⁻²⁵ m, h = 6.625 × 10⁻³⁴ Js.

v = h/(mλ) = 6.625 × 10⁻³⁴ / (1.0 × 10⁻⁹ × 3 × 10⁻²⁵) = 6.625 × 10⁻³⁴ / (3 × 10⁻³⁴) ≈ 2.2 m/s.

✓Final answer
$(d) 2.2 ms⁻¹.$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The de Broglie wavelength (lambda) associated with an electron accelerated through a potential difference of 121 V is ___. [m_e = 9.1 x 10^-31 kg, h = 6.63 x 10^-34 Js]$
(a) 12.0 A
(b) 2.1 A
(c) 1.12 A
(d) 0.12 A
›Reveal solutionSolution

For an electron accelerated through V volts, lambda = 12.27/sqrt(V) angstrom; with V = 121 V this is 1.12 A.

The de Broglie wavelength of an electron accelerated through potential difference V is:

lambda = h/sqrt(2 m e V) = 12.27/sqrt(V) angstrom (a standard result).

lambda = 12.27/sqrt(121) = 12.27/11 = 1.115 A approximately 1.12 A.

✓Final answer
$(c) 1.12 A.$
GUJCET 2019Set 131 markMCQ
$0.5\times10^{-10}$
›Reveal solutionSolution
$\lambda\propto E^{-1/2}$ , so doubling $\lambda$ requires $E$ reduced to one-fourth.

Concept: de Broglie wavelength $\lambda=\dfrac{h}{\sqrt{2mE}}\propto\dfrac{1}{\sqrt E}$ , therefore $E\propto\dfrac{1}{\lambda^2}$ .

Steps:

$\lambda$ increases by factor $\dfrac{10^{-10}}{0.5\times10^{-10}}=2$ .

$E\propto\lambda^{-2}$ , so $E$ changes by $\left(\dfrac1{2}\right)^2=\dfrac14$ .

Energy is decreased to one-fourth part.

✓Final answer
$Option (A) — Decreased to fourth part$

ANSWER: (A)
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. The uncertainty in position of a particle is same as it's de Broglie wavelength, uncertainty in its momentum is ___.$
(a) h/lambda
(b) 2h/3lambda
(c) lambda/h
(d) 3lambda/2h
›Reveal solutionSolution

Using Heisenberg's relation delta_x . delta_p approximately h with delta_x = lambda gives delta_p = h/lambda.

Heisenberg's uncertainty principle (in the simple form): delta_x . delta_p approximately h.

Given the uncertainty in position equals the de Broglie wavelength, delta_x = lambda.

Then delta_p approximately h / delta_x = h / lambda.

(Note: the de Broglie relation itself is lambda = h/p, so this says the momentum uncertainty is comparable to the momentum itself.)

✓Final answer
$(a) h/lambda.$
GUJCET 2015Set C1 markMCQ
$2:1$
›Reveal solutionSolution
[!TLDR]

$\lambda_\alpha = h/(4uv)$ and $\lambda_d = h/(4uv)$ are equal, so $\lambda_\alpha:\lambda_d = 1:1$ . Answer: (C).

Concept

The de Broglie wavelength of a particle is $\lambda = \dfrac{h}{mv}$ (NCERT/CBSE dual nature of matter). An alpha particle has mass $\approx 4u$ ; a deuteron has mass $\approx 2u$ .

Solution

For the alpha particle (mass $4u$ , speed $v$ ):
$\lambda_\alpha = \frac{h}{(4u)(v)} = \frac{h}{4uv}$
For the deuteron (mass $2u$ , speed $2v$ ):
$\lambda_d = \frac{h}{(2u)(2v)} = \frac{h}{4uv}$
The momenta $mv$ are identical, so
$\frac{\lambda_\alpha}{\lambda_d} = \frac{h/4uv}{h/4uv} = 1 \;\Rightarrow\; 1:1$
[!ANSWER]

(C) $1:1$ .
GUJCET 2015Set C1 markMCQ
$\dfrac{h}{\sqrt{3mKT}}$
›Reveal solutionSolution
[!TLDR] $\lambda = h/p$ with $p = \sqrt{3mKT}$ gives $\lambda = \dfrac{h}{\sqrt{3mKT}}$ .

Concept

A particle in thermal equilibrium at temperature $T$ has average translational kinetic energy $KE = \tfrac{3}{2}kT$ (k = Boltzmann constant). Its momentum is $p = \sqrt{2m\,KE}$ , and the de-Broglie wavelength is $\lambda = h/p$ .

Solution

$KE = \dfrac{3}{2}kT$

$p = \sqrt{2m\,KE} = \sqrt{2m\cdot\tfrac{3}{2}kT} = \sqrt{3mkT}$

$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{3mKT}}$ .

This matches option A. (Options C and D are also dimensionally wrong for a wavelength.)

[!ANSWER] (A)
GUJCET 2014Set A1 markMCQ
$\sqrt{2}$
›Reveal solutionSolution
[!TLDR]

The new wavelength is $\dfrac{1}{\sqrt{2}}$ times the original.

Concept

For a particle of mass $m$ and kinetic energy $E$ , momentum $p = \sqrt{2mE}$ , so the de Broglie wavelength is
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{E}}.$
Solution

If $E \to 2E$ (mass unchanged):
$\frac{\lambda_{new}}{\lambda} = \sqrt{\frac{E}{2E}} = \frac{1}{\sqrt{2}}.$
So $\lambda_{new} = \dfrac{\lambda}{\sqrt{2}}$ .

[!ANSWER]

(B)

←Previous question11.37 — Answer the following questions: (a) Quarks inside protons and neutrons…→Next questionQ2 — The wavelength of a photon needed to remove a proton from a nucleus wh…

Q.A particle is dropped from a height HHH. The de Broglie wavelength of the particle as a function of height is proportional to

🎓Unlock everything free for 14 days

Q.A particle is dropped from a height $H$ . The de Broglie wavelength of the particle as a function of height is proportional to

Unlock everything free for 14 days