If the position of the electron is measured within an accuracy of $\pm 0.002$ nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $\dfrac{h}{4\pi m} \times 0.05$ nm, is there any problem in defining this value?

If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the

Exercises · 2.61

\pm 0.002

Sikkim CbseNCERTSubjective· 2mImportance★★★★★

56% · 79/140 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

From Heisenberg's principle with $\Delta x = 0.002\ \text{nm}$ , the uncertainty in momentum is $\Delta p = 2.64 \times 10^{-23}\ \text{kg m s}^{-1}$ . The quoted momentum is smaller than this uncertainty, so it cannot be defined.

Heisenberg uncertainty principle:

\Delta x\,\Delta p \ge \frac{h}{4\pi}

With $\Delta x = 0.002\ \text{nm} = 2 \times 10^{-12}\ \text{m}$ and $h = 6.626 \times 10^{-34}\ \text{J s}$ , the minimum uncertainty in momentum is:

\Delta p = \frac{h}{4\pi\,\Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 2 \times 10^{-12}} = 2.64 \times 10^{-23}\ \text{kg m s}^{-1}

Assessing the quoted momentum. As printed, $\dfrac{h}{4\pi m}\times 0.05\ \text{nm}$ is dimensionally inconsistent (it does not have units of momentum, $\text{kg m s}^{-1}$ ), so the expression contains a typographical defect. Taking the intended form $p = \dfrac{h}{4\pi \times 0.05\ \text{nm}}$ :

$p = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.05 \times 10^{-9}} = 1.05 \times 10^{-23}\ \text{kg m s}^{-1}$ …

Start your 14-day free trial to unlock the full solution →

←Previous question · exercises2.60 — The velocity associated with a proton moving in a potential difference…→Next question · exercises2.62 — The quantum numbers of six electrons are given below. Arrange them in …

Q.If the position of the electron is measured within an accuracy of ±0.002\pm 0.002±0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h4πm×0.05\dfrac{h}{4\pi m} \times 0.054πmh​×0.05 nm, is there any problem in defining this value?

🎓Unlock everything free for 14 days

Q.If the position of the electron is measured within an accuracy of $\pm 0.002$ nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $\dfrac{h}{4\pi m} \times 0.05$ nm, is there any problem in defining this value?

Unlock everything free for 14 days