An electron is moving with an initial velocity $\vec{v} = v_0\,\hat{i}$ and is in a magnetic field $\vec{B} = B_0\,\hat{j}$. Then its de Broglie wavelength (a) remains constant. (b) increases with time. (c) decreases with time. (d) increases and decreases periodically.

> [!TLDR] > The electron moves in a circular path due to the magnetic force, so its speed stays constant. Since de Broglie wavelength depends only on speed, it remains constant. The correct option is **(A)**. The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$, where $p$ is the magnitude of its momentum. For an electron, $p = m v$ (non-relativistic here), so $\lambda = \frac{h}{m v}$. The key question is: does the speed $v$ of the electron change? Let’s see what happens when a charged particle enters a magnetic field. 1. **The magnetic force does no work.** The Lorentz force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is $\vec{F} = q (\vec{v} \times \vec{B})$. This force is always perpendicular to $\vec{v}$. Since power $P = \vec{F} \cdot \vec{v} = 0$, the kinetic energy $K = \frac{1}{2} m v^2$ remains constant. Therefore, the speed $v$ does **not** change. 2. **The path is circular in the plane perpendicular to $\vec{B}$.** Here $\vec{v} = v_0 \hat{i}$ and $\vec{B} = B_0 \hat{j}$. The cross product $\vec{v} \times \vec{B} = v_0 B_0 (\hat{i} \times \hat{j}) = v_0 B_0 \hat{k}$. So the force is along $\hat{k}$ (the $z$-direction), perpendicular to both velocity and field. This force provides the centripetal acceleration, bending the electron into a circle in the $xz$-plane. The speed remains $v_0$ throughout. …

An electron is moving with an initial velocity vecv = v₀hati and is in a magnetic field vecB = B₀hatj. Then it

NCERT Exemplar · Q5

\vec{v} = v_0\,\hat{i}

(a) remains constant.

(b) increases with time.

(d) increases and decreases periodically.

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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 moves in a circular path due to the magnetic force, so its speed stays constant. Since de Broglie wavelength depends only on speed, it remains constant. The correct option is (A).

The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$ , where $p$ is the magnitude of its momentum. For an electron, $p = m v$ (non-relativistic here), so $\lambda = \frac{h}{m v}$ . The key question is: does the speed $v$ of the electron change?

Let’s see what happens when a charged particle enters a magnetic field.

The magnetic force does no work.

The Lorentz force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is $\vec{F} = q (\vec{v} \times \vec{B})$ . This force is always perpendicular to $\vec{v}$ .

Since power $P = \vec{F} \cdot \vec{v} = 0$ , the kinetic energy $K = \frac{1}{2} m v^2$ remains constant. Therefore, the speed $v$ does not change.
The path is circular in the plane perpendicular to $\vec{B}$ .

Here $\vec{v} = v_0 \hat{i}$ and $\vec{B} = B_0 \hat{j}$ . The cross product $\vec{v} \times \vec{B} = v_0 B_0 (\hat{i} \times \hat{j}) = v_0 B_0 \hat{k}$ . So the force is along $\hat{k}$ (the $z$ -direction), perpendicular to both velocity and field.

This force provides the centripetal acceleration, bending the electron into a circle in the $xz$ -plane. The speed remains $v_0$ throughout. …

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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

$\lambda'=\frac{h}{p}=\frac{h}{h/\lambda}=\lambda$ …
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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.

…

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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. …

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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.

…

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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: …

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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 $\dfrac{10^{-10}}{0.5\times10^{-10}}=2$ . …
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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.

…

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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 $v$ ):

$\lambda_\alpha = \frac{h}{(4u)(v)} = \frac{h}{4uv}$ …
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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

$KE = \dfrac{3}{2}kT$ …
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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): …
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Q.An electron is moving with an initial velocity $\vec{v} = v_0\,\hat{i}$ and is in a magnetic field $\vec{B} = B_0\,\hat{j}$ . Then its de Broglie wavelength

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)

Method: Deciding Whether a de Broglie Wavelength Changes Under a Given Force

Steps

Concept

Solution

Concept

Solution

Concept

Solution

Q.An electron is moving with an initial velocity v⃗=v0 i^\vec{v} = v_0\,\hat{i}v=v0​i^ and is in a magnetic field B⃗=B0 j^\vec{B} = B_0\,\hat{j}B=B0​j^​. Then its de Broglie wavelength

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Q.An electron is moving with an initial velocity $\vec{v} = v_0\,\hat{i}$ and is in a magnetic field $\vec{B} = B_0\,\hat{j}$ . Then its de Broglie wavelength

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