Before You Begin

Question #95969

Difficulty: 5/10 | 🟢 Free

The Question

A photoelectric emitter (work function $\Phi$ ) and a collector are configured as large parallel plates separated by a distance $d$ . The entire setup is immersed in a transparent liquid dielectric of relative permittivity $\epsilon_r > 1$ . Light of wavelength $\lambda$ is incident on the emitter. To stop the photoelectrons from reaching the collector, the plates are electrically isolated and given a free surface charge density of magnitude $\sigma$ (positive on the emitter, negative on the collector). Due to the polarization of the medium, the effective potential difference doing work on the electrons is screened such that $V_{eff} = V_{vac} / \epsilon_r$ , where $V_{vac}$ is the potential difference the same free charge would produce in vacuum. What is the minimum free surface charge density $\sigma$ required to completely halt the photoelectrons?

Options

A.
$\frac{\epsilon_0}{e d} \left( \frac{hc}{\lambda} - \Phi \right)$
B.
$\frac{\epsilon_r \epsilon_0}{e d} \left( \frac{hc}{\lambda} - \Phi \right)$
C.
$\frac{\epsilon_0}{\epsilon_r e d} \left( \frac{hc}{\lambda} - \Phi \right)$
D.
$\frac{\epsilon_r \epsilon_0}{e d} \left( \frac{hc}{\lambda} + \Phi \right)$

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

Stopping Potential Sign|Potential of Concentric Shells|Equilibrium Stability|Approximations

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the intersection of quantum photoemission and macroscopic electrostatics in linear dielectrics.

The 'Lethal Collision' occurs when students treat the stopping potential purely as a property of the metal and incident light ( $V_s = K_{max}/e$ ), directly equating it to the vacuum formula for the electric field of parallel plates ( $\sigma d / \epsilon_0$ ).

They fail to account for the medium's polarization, which screens the free charge and reduces the effective macroscopic electric field doing work on the electron ( $E = \sigma / \epsilon_r \epsilon_0$ ).

Consequently, to achieve the same required stopping potential difference, the external free surface charge density must be scaled up by a factor of the relative permittivity ( $\epsilon_r$ ).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Dual Nature of Radiation and Matter - Class 12
Concept	Stopping Potential Sign
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free