⇦ | ⇨ |
The mean free path of electrons in a metal is 4 x 10⁻⁸ m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of V/m
Options
(a) 5 x 10⁻¹¹
(b) 8 x 10⁻¹¹
(c) 5 x 10⁷
(d) 8 x 10⁷
Correct Answer:
5 x 10⁷
Explanation:
E = V / d = 2 / 4 x 10⁻⁸
0.5 x 10⁸ = 5 x 10⁷ Vm⁻¹
Related Questions: - An observer moving away from a stationary source. The observed frequency is half
- Two bulbs 40 watt and 60 watt and rated voltage 240 V are connected in series
- An electron in the hydrogen atom jumps from excited state n to the ground state
- Two parallel beams of positron moving in the same direction will
- Two wheels having radii in the ratio 1:3 are connected by a common belt
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An observer moving away from a stationary source. The observed frequency is half
- Two bulbs 40 watt and 60 watt and rated voltage 240 V are connected in series
- An electron in the hydrogen atom jumps from excited state n to the ground state
- Two parallel beams of positron moving in the same direction will
- Two wheels having radii in the ratio 1:3 are connected by a common belt
Topics: Electrostatics (146)
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
Leave a Reply