Physics
JAMB 2000
Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is
-
A.
1.7V
-
B.
2.0V
-
C.
8.0V
-
D.
15.0V
Correct Answer: Option B
Explanation
For a photoemitted electron,
eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V
Report an Error
Ask A Question
Download App
Quick Questions
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}