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Question #43589

Difficulty: 8/10 | 🟢 Free

The Question

An electron (rest mass $m_0$ , charge magnitude $e$ ) is accelerated from rest through a potential difference $V = \frac{4m_0 c^2}{e}$ , where $c$ is the speed of light in vacuum. The accelerated electron beam is incident normally on a single slit of width $d$ to produce a Fraunhofer diffraction pattern on a screen at a distance $D$ ( $D \gg d$ ). Let $W_r$ be the true width of the central maximum calculated using relativistic mechanics, and $W_c$ be the width calculated assuming classical (non-relativistic) mechanics. What is the ratio $\frac{W_r}{W_c}$ ?

Options

A.
$\frac{1}{\sqrt{3}}$
B.
$\frac{1}{\sqrt{5}}$
C.
$\sqrt{3}$
D.
$\frac{2}{3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Diffraction Width|Davisson-Germer Significance|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the transition from classical to relativistic regimes in quantum mechanics (wave-particle duality).

Students often reflexively apply the classical momentum-kinetic energy relation ( $p = \sqrt{2mK}$ ) to find the de Broglie wavelength, regardless of the accelerating potential.

When the kinetic energy $eV$ is comparable to or exceeds the rest mass energy $m_0 c^2$ , the relativistic energy-momentum relation ( $E^2 = p^2 c^2 + m_0^2 c^4$ ) must be used.

The trap lies in failing to recognize that the classical formula underestimates the momentum of highly relativistic electrons, thereby overestimating the de Broglie wavelength and the resulting diffraction width.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Diffraction Width
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free