Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}mv^2$ becomes comparable with $mc^2$, where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron? (a) $\lambda = 10\ \text{nm}$ (b) $\lambda = 10^{-1}\ \text{nm}$ (c) $\lambda = 10^{-4}\ \text{nm}$ (d) $\lambda = 10^{-6}\ \text{nm}$

Relativistic corrections become necessary when the expression for the kinetic energy (1)/(2)mv² becomes compar

NCERT Exemplar · Q24

\frac{1}{2}mv^2

(a)

\lambda = 10\ \text{nm}

(b)

\lambda = 10^{-1}\ \text{nm}

(c)

\lambda = 10^{-4}\ \text{nm}

(d)

\lambda = 10^{-6}\ \text{nm}

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Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

You already know that light behaves like a wave (interference, diffraction) and like a particle (photoelectric effect). That's wave-particle duality for light. De Broglie's radical idea in 1924 was: if light can be both, why can't matter be both too?

He proposed that every moving particle — an electron, a proton, even a cricket ball — has a wavelength associated with it. The faster it moves, the shorter that wavelength becomes.

The Intuition

Think of a wave on a string. Its wavelength is the distance between two consecutive crests. Now imagine an electron moving through space. De Broglie said that the electron's motion itself creates a "matter wave" — a wave of probability that guides where the electron is likely to be found.

Note

You never see this wavelength in everyday life because for large objects it's unimaginably tiny. A cricket ball moving at 30 m/s has a de Broglie wavelength of about $10^{-34}$ m — far smaller than an atomic nucleus. That's why macroscopic objects behave like particles.

The Precise Statement

The de Broglie wavelength $\lambda$ of a particle is given by:

\lambda = \frac{h}{p}

where:

$h$ is Planck's constant ( $6.626 \times 10^{-34} \ \text{J·s}$ )
$p$ is the momentum of the particle ( $p = mv$ for non-relativistic speeds)

Key point: The wavelength depends only on momentum, not on charge, mass, or any other property. A fast electron and a slow proton can have the same wavelength if their momenta are equal.

What This Means Physically

For electrons in atoms: The de Broglie wavelength of an electron in a hydrogen atom is roughly the size of the atom itself ( $\approx 10^{-10}$ m). This is why electrons form standing waves around the nucleus — only certain wavelengths "fit" into the orbit, which explains quantised energy levels.
For experiments: If you fire electrons through a crystal, they diffract just like X-rays. This was confirmed by Davisson and Germer in 1927 — a Nobel-winning experiment that proved de Broglie right.
For large objects: The wavelength is so small that wave behaviour is undetectable. A car moving at 100 km/h has $\lambda \approx 10^{-38}$ m — you'd need a slit smaller than an atom to see diffraction.

Watch out

A common mistake is to think the de Broglie wavelength is the size of the particle. It is not. It is the wavelength of the probability wave associated with the particle. The particle itself remains point-like.

Worked Example

Question: What is the de Broglie wavelength of an electron moving at $2.0 \times 10^6$ m/s? (Mass of electron $m_e = 9.11 \times 10^{-31}$ kg)

Solution:

First, find momentum:

p = mv = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \ \text{kg·m/s}

Then apply de Broglie's formula: …

Why this formula?

De Broglie Wavelength: Why Matter Has a Wave Nature

The idea that a moving particle has a wavelength is one of the most radical shifts in physics. It came from Louis de Broglie in 1924, who asked a simple question: if light — which we thought was a wave — can behave like a particle (the photon), then why can't a particle behave like a wave?

The Core Insight: Symmetry in Nature

De Broglie started from Einstein's relation for a photon. For light, the energy $E$ and momentum $p$ of a photon are linked to its wave properties — frequency $f$ and wavelength $\lambda$ — by:

E = h f \quad \text{and} \quad p = \frac{h}{\lambda}

where $h$ is Planck's constant. These are not arbitrary; they come from the fact that light is an electromagnetic wave, and Planck had already shown that energy comes in quanta $hf$ .

De Broglie's reasoning was a leap of symmetry: if nature treats light and matter on equal footing (as Einstein's special relativity suggests), then any moving particle should also have a wavelength associated with it. He proposed that the same relation holds for matter:

\lambda = \frac{h}{p}

where $p = mv$ is the momentum of the particle (for non-relativistic speeds). This is the de Broglie wavelength.

Why This Formula Makes Sense: A Simple Derivation

There is no rigorous "derivation" from first principles — de Broglie's hypothesis was a postulate. But we can see why it is plausible by combining two key ideas from relativity and quantum theory.

Step 1: Energy of a particle from relativity

For a particle with rest mass $m_0$ , the total energy in special relativity is:

E = \sqrt{p^2 c^2 + m_0^2 c^4}

For a photon, $m_0 = 0$ , so $E = pc$ . This matches the photon's wave relation $E = hf$ and $p = h/\lambda$ .

Step 2: Assume the same wave-particle duality for matter

If a massive particle also has a wave associated with it, then its energy should also be $E = hf$ , where $f$ is the frequency of the matter wave. Equating the relativistic energy with the quantum energy:

hf = \sqrt{p^2 c^2 + m_0^2 c^4}

For a particle moving at non-relativistic speeds ( $v \ll c$ ), the momentum $p = mv$ is small compared to $m_0 c$ , so we can expand:

E \approx m_0 c^2 + \frac{p^2}{2m_0}

The first term is rest energy, which is constant. The second term is kinetic energy $K = p^2/(2m)$ . The wave frequency $f$ then corresponds to the kinetic part (since rest energy doesn't contribute to motion). But the key relation we want is between wavelength and momentum.

Step 3: The wavelength from the wave speed

For any wave, the speed $v_{\text{wave}} = f \lambda$ . For a matter wave, de Broglie proposed that the wave speed equals the particle's speed $v$ (this is the phase velocity). So:

v = f \lambda

Now use $E = hf$ and $E = \frac{1}{2}mv^2$ (non-relativistic kinetic energy). Then:

f = \frac{E}{h} = \frac{mv^2}{2h}

Substitute into $v = f \lambda$ :

v = \frac{mv^2}{2h} \lambda \quad \Rightarrow \quad \lambda = \frac{2h}{mv}

This gives $\lambda = 2h/p$ , which is wrong by a factor of 2. The correct formula is $\lambda = h/p$ . …

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Relativistic corrections set in when $\tfrac{1}{2}mv^2\sim mc^2$ , i.e. when the de Broglie wavelength falls to about the electron's Compton wavelength ( $\sim 10^{-3}$ nm). Among the options the picometre-range choice is (C) $10^{-4}$ nm.

When do relativistic corrections matter?

Newtonian kinetic energy $\tfrac{1}{2}mv^2$ is a good approximation only while it is small compared with the rest energy $mc^2$ . Corrections become important when the two are comparable:

\frac{1}{2}mv^2 \sim mc^2.

Turn the condition into a wavelength

Write the kinetic energy through the momentum, $K=\dfrac{p^2}{2m}$ , and set it comparable to $mc^2$ :

\frac{p^2}{2m}\sim mc^2 \quad\Rightarrow\quad p\sim\sqrt{2}\,mc.

The de Broglie wavelength at this momentum is

\lambda = \frac{h}{p} \sim \frac{h}{\sqrt{2}\,mc},

which is essentially the electron's Compton wavelength $\lambda_C=\dfrac{h}{mc}$ (up to the factor $\sqrt{2}$ ).

Put in the numbers

$\lambda \sim \frac{6.63\times10^{-34}}{\sqrt{2}\,(9.11\times10^{-31})(3\times10^{8})} \approx 1.7\times10^{-12}\ \text{m} = 1.7\times10^{-3}\ \text{nm}.$ …

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

Showing the 12 most recent of 16 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$1 \times 10^{-31}$
›Reveal solutionSolution

Get momentum from $p = h/\lambda$ , then kinetic energy from $p^2/2m$ : the result is $5\times10^{-22}$ J.

Momentum:
$p = \frac{h}{\lambda} = \frac{6.3\times10^{-34}}{63\times10^{-9}} = \frac{6.3\times10^{-34}}{6.3\times10^{-8}} = 1\times10^{-26}\ \text{kg m s}^{-1}.$
Kinetic energy: …
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KEAM 2026Set eng-2026-04184 marksMCQ
$Q. For the same kinetic energy, the de Broglie wavelengths associated with particles of different masses are (A) directly proportional to their masses (B) directly proportional to the square root of their masses (C) inversely proportional to the square root of their masses (D) inversely proportional to their masses (E) directly proportional to the square of their masses$
›Reveal solutionSolution

The de Broglie wavelength $\lambda=h/\sqrt{2mE}$ , so at fixed kinetic energy it is inversely proportional to $\sqrt m$ .

de Broglie wavelength in terms of kinetic energy $E$ :
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$
For the same kinetic energy $E$ , $h$ and $E$ are constant, so …
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KEAM 2026Set eng-2026-04204 marksMCQ
$\lambda_e$
›Reveal solutionSolution
[!TLDR]

At equal kinetic energy the de Broglie wavelength scales as $1/\sqrt{m}$ , so the heavier proton has the shorter wavelength and $\lambda_p:\lambda_e = 1:\sqrt{1836}$ .

Concept

The de Broglie wavelength is $\lambda = h/p$ . For a non-relativistic particle the momentum relates to kinetic energy $E$ by $p = \sqrt{2mE}$ , so $\lambda = h/\sqrt{2mE}$ . This wave-nature-of-matter relation is part of the NCERT/CBSE dual-nature chapter.

Solution

For equal kinetic energies $E$ ,
$\lambda = \frac{h}{\sqrt{2mE}} \implies \lambda \propto \frac{1}{\sqrt{m}}.$
Taking the ratio for the proton and electron, …
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KEAM 2026Set eng-2026-04214 marksMCQ
$0.1227\,nm$
›Reveal solutionSolution

de Broglie relation for electrons: $\lambda=12.27/\sqrt{V}\\,$ \u00c5 gives $V=100\\,$ V.

For an electron accelerated through $V$ volts,

$\lambda = \frac{12.27}{\sqrt{V}}\\,\text{\u00c5}.$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\sqrt{3}:\sqrt{2}:1$
›Reveal solutionSolution

de Broglie: $\lambda=h/p$ , so $p\propto 1/\lambda$ . Inverting $1:2:3$ gives $6:3:2$ .

From $\lambda=\dfrac{h}{p}$ , momentum is inversely proportional to wavelength: …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\lambda$
›Reveal solutionSolution
The de-Broglie wavelength of a photon equals the wavelength lambda of its own radiation.

Concept and Intuition

A photon of electromagnetic wavelength lambda carries momentum p = h/lambda. Its de-Broglie wavelength is defined as lambda_dB = h/p, which returns exactly lambda. So the two coincide for a photon.

Step-by-Step Solution

Photon momentum p = h/lambda.

de-Broglie wavelength = h/p = h/(h/lambda) = lambda. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$6.63 \times 10^{-31}$
›Reveal solutionSolution

The de Broglie wavelength $\lambda = h/(mv)$ works out to $10^{-9}$ m, i.e. 1.0 nm.

The de Broglie relation gives

$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-31})(1 \times 10^6)}$ …
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KEAM 2025Set eng-2025-04274 marksMCQ
$1:1$
›Reveal solutionSolution

Since $\lambda \propto \dfrac{1}{\sqrt{K}}$ for equal masses, the wavelength ratio is $\sqrt{K_2/K_1} = \sqrt{2/0.02} = 10:1$ .

The de Broglie wavelength in terms of kinetic energy $K$ is:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}.$
For particles of equal mass, $\lambda \propto \dfrac{1}{\sqrt{K}}$ . Taking $K_1 = 0.02\ \text{eV}$ and $K_2 = 2\ \text{eV}$ : …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$2.2 \times 10^{-2}\ kg$
›Reveal solutionSolution
[!TLDR]

Applying de Broglie's relation $\lambda = h/mv$ , the numerical coefficients divide to exactly 1 and the scale works out to picometres, giving $\lambda = 1$ pm.

Concept

Louis de Broglie proposed that every moving particle has an associated wavelength $\lambda = \dfrac{h}{mv}$ , where $h$ is Planck's constant and $mv$ is the momentum. This wave-particle duality is a core idea in the NCERT/CBSE-aligned 'Structure of Atom' chemistry syllabus that KEAM follows.

Solution
$\lambda = \frac{h}{mv} = \frac{6.6\times 10^{-34}}{(2.2\times 10^{-27})(3.0\times 10^{5})}.$
The coefficients divide cleanly:
$\frac{6.6}{2.2\times 3.0} = \frac{6.6}{6.6} = 1.$
The powers of ten give
$\frac{10^{-34}}{10^{-27}\cdot 10^{5}} = 10^{-34+27-5} = 10^{-12}\text{ m}.$
Hence

$\lambda = 1\times 10^{-12}\text{ m} = 1\text{ pm}.$ …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$\left(2 \times 10^{10}\right)h\ kgms^{-1}$
›Reveal solutionSolution

The de Broglie wavelength $\lambda = h/p$ ; here $p = (2\times10^{10})h$ , so $\lambda = 1/(2\times10^{10})\,\mathrm{m} = 0.5\,\text{\AA}$ .

The de Broglie wavelength is
$\lambda = \frac{h}{p}.$
Given $p = (2\times10^{10})\,h\ \mathrm{kg\,m\,s^{-1}}$ , the Planck constant cancels: …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\gamma$
›Reveal solutionSolution

$\lambda \approx 1.36$ Å for $81$ V electrons $\Rightarrow$ X-ray region.

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

$\lambda = \frac{1.227}{\sqrt{V}}\ \text{nm} = \frac{1.227}{\sqrt{81}} = \frac{1.227}{9} = 0.136\ \text{nm} = 1.36\ \text{\AA}.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$\frac{1}{3000}$
›Reveal solutionSolution

$\lambda=h/mv$ , so the ratio is $1/(2000\times3)=1/6000$ .

The de Broglie wavelength is $\lambda=\dfrac{h}{mv}$ , inversely proportional to the product $mv$ . …
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Q.Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}mv^2$ becomes comparable with $mc^2$ , where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Intuition

The Precise Statement

What This Means Physically

Worked Example

Why this formula?

De Broglie Wavelength: Why Matter Has a Wave Nature

The Core Insight: Symmetry in Nature

Why This Formula Makes Sense: A Simple Derivation

When do relativistic corrections matter?

Turn the condition into a wavelength

Put in the numbers

Method: Estimating an Order-of-Magnitude Threshold from a Physical Condition

Steps

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Concept

Solution

Q.Relativistic corrections become necessary when the expression for the kinetic energy 12mv2\frac{1}{2}mv^221​mv2 becomes comparable with mc2mc^2mc2, where mmm is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

When do relativistic corrections matter?

Turn the condition into a wavelength

Put in the numbers

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Q.Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}mv^2$ becomes comparable with $mc^2$ , where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

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