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.

CBSE 2024Set ANNUAL1 mark
$Q. What is the ratio of de-Broglie wavelength associated with two electron beams A and B accelerated through 25 V and 36 V respectively ?$
›Reveal solutionSolution

de Broglie wavelength for an electron accelerated through potential V: $\lambda \propto 1/\sqrt V$ .

For an electron accelerated through a potential difference V, $\lambda = \dfrac{h}{\sqrt{2meV}}$ , so $\lambda \propto \dfrac{1}{\sqrt V}$ .

…

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CBSE 2019Set 55/5/11 mark

Q. Plot a graph of the de-Broglie wavelength associated with a proton versus its momentum.

›Reveal solution

The de-Broglie wavelength $\lambda$ of a proton is inversely proportional to its momentum $p$ , following $\lambda = h/p$ . The graph is a rectangular hyperbola in the first quadrant — a smooth curve that falls steeply at low momentum and flattens as momentum increases.

The idea is beautifully simple. Louis de Broglie proposed that every moving particle has a wavelength associated with it, given by $\lambda = h/p$ , where $h$ is Planck’s constant and $p$ is the linear momentum. For a proton, this relation holds exactly — no approximations, no mass dependence once you write it in terms of $p$ .

Why does this matter? Because the graph tells you something immediate: as you speed up a proton (increase its momentum), its wavelength shrinks. At low speeds, the wavelength is large; at very high speeds, it becomes tiny. The shape is a rectangular hyperbola — the same curve you get for $xy = \text{constant}$ .

Let’s build the graph step by step.

Write the relation. The de-Broglie wavelength is

\lambda = \frac{h}{p}

where $h = 6.63 \times 10^{-34} \ \text{J·s}$ . For a proton, $p = mv$ (non-relativistic) or $p = \gamma mv$ (relativistic), but the $\lambda$ vs $p$ relation itself is universal — it doesn’t care about mass or speed.

Identify the axes.
- Horizontal axis: momentum $p$ (in kg·m/s, say).
- Vertical axis: wavelength $\lambda$ (in metres). Both are positive quantities, so the graph lies entirely in the first quadrant.
Understand the shape.

The equation $\lambda = h/p$ is of the form $y = c/x$ , a rectangular hyperbola. Key features:
- As $p \to 0^+$ , $\lambda \to \infty$ — the curve shoots up near the vertical axis.
- As $p \to \infty$ , $\lambda \to 0$ — the curve approaches the horizontal axis asymptotically.
- The product $\lambda p = h$ is constant, so every point on the curve satisfies this.

Plot a few points to guide the sketch.

Let’s pick some convenient momentum values and compute $\lambda$ :

$p$ (kg·m/s)	$\lambda = h/p$ (m)
$1 \times 10^{-24}$	$6.63 \times 10^{-10}$
$2 \times 10^{-24}$	$3.32 \times 10^{-10}$
$5 \times 10^{-24}$	$1.33 \times 10^{-10}$
$1 \times 10^{-23}$	$6.63 \times 10^{-11}$

Notice: doubling $p$ halves $\lambda$ — that’s the inverse proportionality in action. …

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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

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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