| ⇦ | 
| ⇨ |
A Zenor diode has a contact potential of 1 V in the absence of biasing. It undergoes Zenor breakdown for an electric field of 10⁶ V/m at the depletion region of p-n junction. If the width of the depletion region is 2.5 μm, what should be the reverse biased potential for the Zenor breakdown to occur?
Options
(a) 3.5 V
(b) 2.5 V
(c) 1.5 V
(d) 0.5 V
Correct Answer:
2.5 V
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A student measures the terminal potential difference (V) of a cell
- Two small spherical shells A and B are given positive charges of 9C and 4C
- If critical angle for a material to air passage is 30°, the refractive index
- A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest
- A cricketer catches a ball of mass 150 g in 0.1 sec moving with speed 20 m/s
Topics: Electronic Devices
(124)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A student measures the terminal potential difference (V) of a cell
- Two small spherical shells A and B are given positive charges of 9C and 4C
- If critical angle for a material to air passage is 30°, the refractive index
- A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest
- A cricketer catches a ball of mass 150 g in 0.1 sec moving with speed 20 m/s
Topics: Electronic Devices (124)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
V=E*d
E=10^6 v/m
d=2.5 μm
Therefor V=2.5