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Question #43620

Difficulty: 8/10 | 🟢 Free

The Question

An isolated parallel-plate capacitor in vacuum carries a constant free charge, creating a potential difference $V$ between its plates. An electron of mass $m$ and charge $-e$ is released from rest from the negative plate. It accelerates across the gap and reaches the positive plate with a de Broglie wavelength $\lambda_0$ . The space between the plates is then completely filled with a linear, homogeneous dielectric liquid of dielectric constant $K$ . The free charge on the plates remains unchanged. If an identical electron is now released from rest from the negative plate, what is its final de Broglie wavelength just before reaching the positive plate? (Assume the dielectric offers no viscous drag and does not alter the electron's rest mass).

Options

A.
$\lambda_0 \sqrt{K}$
B.
$\frac{\lambda_0}{\sqrt{K}}$
C.
$\lambda_0 K$
D.
$\lambda_0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

De Broglie Wavelength Scaling|Dielectric Constant (K) Shift|Electrostatic Potential and Capacitance

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of electrostatics in linear media and quantum mechanics.

The primary trap is the assumption that the potential difference remains $V$ (confusing an isolated capacitor with a battery-connected one), which would lead students to incorrectly choose option D ( $\lambda_0$ ).

Another common mistake is misapplying the dielectric constant directly to the wavelength formula without routing it through the Work-Energy theorem, leading to inverted or un-rooted factors (Options B and C).

The core physical misconception is failing to recognize that in an isolated system, dielectric polarization creates a shielding field that reduces the net electric field to $E_0/K$ .

This reduces the actual work done on the electron to $eV/K$ , resulting in a lower kinetic energy and a correspondingly longer de Broglie wavelength.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Dual Nature of Radiation and Matter - Class 12
Concept	De Broglie Wavelength Scaling
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free