Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneous

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Q. Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously. Reason (R): The path of an electron in an atom is clearly defined.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true and R is not the correct explanation of A.

(iii) A is true and R is false.

(iv) Both A and R are false.

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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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Assertion (A) correctly states Heisenberg's Uncertainty Principle, making it true. Reason (R) incorrectly describes electron behavior in an atom, making it false. Therefore, A is true and R is false.

The question asks us to evaluate an Assertion (A) and a Reason (R) related to the behavior of electrons in an atom. This requires understanding fundamental principles of quantum mechanics, specifically Heisenberg's Uncertainty Principle and the quantum mechanical model of the atom.

Concept and Intuition: Heisenberg's Uncertainty Principle

In the quantum world, particles like electrons exhibit wave-particle duality. This means they don't behave purely as tiny, hard spheres with definite positions and momenta, nor purely as waves spread out in space. This dual nature leads to some profound consequences, one of which is Heisenberg's Uncertainty Principle.

Imagine trying to precisely locate a wave. If you want to know its exact position, you'd try to pinpoint a very small region where it exists. But to do that, you'd need to combine many different wavelengths, which makes the wave's momentum (related to its wavelength) very uncertain. Conversely, if you want to know the wave's momentum very precisely (meaning it has a very definite wavelength), the wave itself becomes very spread out, making its position highly uncertain.

This isn't a limitation of our measuring instruments; it's a fundamental property of nature. For an electron, we cannot simultaneously know both its exact position and its exact momentum. The more precisely we try to determine one, the less precisely we can know the other.

Step-by-Step Evaluation

Evaluate Assertion (A): "It is impossible to determine the exact position and exact momentum of an electron simultaneously."
- This statement is a direct articulation of Heisenberg's Uncertainty Principle.
- The principle states that for any pair of conjugate variables, such as position ( $x$ ) and momentum ( $p_x$ ), the product of the uncertainties in their simultaneous measurement must be greater than or equal to a certain minimum value. Mathematically, this is expressed as: $\Delta x \Delta p_x \ge \frac{\hbar}{2}$
- This inequality implies that if we try to make $\Delta x$ (uncertainty in position) very small, then $\Delta p_x$ (uncertainty in momentum) must become very large, and vice-versa. It is fundamentally impossible to have both $\Delta x = 0$ and $\Delta p_x = 0$ simultaneously.
- Therefore, Assertion (A) is True.
Evaluate Reason (R): "The path of an electron in an atom is clearly defined." …

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Q.Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously. Reason (R): The path of an electron in an atom is clearly defined.

Concept and Intuition: Heisenberg's Uncertainty Principle

Step-by-Step Evaluation

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Q.Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously.
Reason (R): The path of an electron in an atom is clearly defined.

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