If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6}\ ms^{-1}$, calculate the de Broglie wavelength associated with it.

If the velocity of the electron in Bohr's first orbit is 2.19 × 10⁶ms⁻¹, calculate the de Broglie wavelength a

Exercises · 2.59

2.19 \times 10^{6}\ ms^{-1}

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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 de Broglie wavelength of an electron in Bohr’s first orbit is found by applying $\lambda = h / p$ using the given velocity. The result is $\mathbf{3.32 \times 10^{-10}\ m}$ (or $0.332\ \text{nm}$ ).

The idea here is beautifully simple. De Broglie proposed that every moving particle has a wavelength associated with it — not just light, but matter too. For an electron, which has mass and velocity, its momentum $p = mv$ determines its wavelength through the same relation that works for photons: $\lambda = h / p$ .

In Bohr’s model, the electron in the first orbit has a well-defined velocity. We are given that velocity directly, so we don’t need to derive it from Bohr’s postulates — we just plug into de Broglie’s equation.

\lambda = \frac{h}{mv}

Where:

$h = 6.626 \times 10^{-34}\ \text{J s}$ (Planck’s constant)
$m = 9.1 \times 10^{-31}\ \text{kg}$ (mass of electron)
$v = 2.19 \times 10^{6}\ \text{m/s}$ (given)

Write down the de Broglie relation

\lambda = \frac{h}{mv}

Substitute the values

\lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.19 \times 10^{6})}

First compute the denominator Multiply mass and velocity:

9.1 \times 10^{-31} \times 2.19 \times 10^{6} = 9.1 \times 2.19 \times 10^{-25}

= 19.929 \times 10^{-25} = 1.9929 \times 10^{-24}\ \text{kg m/s}

Now divide

\lambda = \frac{6.626 \times 10^{-34}}{1.9929 \times 10^{-24}}

= \frac{6.626}{1.9929} \times 10^{-10}

$\approx 3.324 \times 10^{-10}\ \text{m}$ …

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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): …
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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

$\lambda=\frac{h}{p}=\frac{h}{mv},$ …
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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. …

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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 …
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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: …
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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 …
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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

$\lambda = \frac{h}{mv}\;\Rightarrow\; \lambda \propto \frac{1}{v}.$ …
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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 $p_\alpha=\dfrac{p_p}{2}$ . …
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Q.If the velocity of the electron in Bohr's first orbit is 2.19×106 ms−12.19 \times 10^{6}\ ms^{-1}2.19×106 ms−1, calculate the de Broglie wavelength associated with it.

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Q.If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6}\ ms^{-1}$ , calculate the de Broglie wavelength associated with it.

Unlock everything free for 14 days