| ⇦ |
| ⇨ |
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: - If the highest modulating frequency of the wave is 5 kHz, the number
- A 6 volt battery is connected to the terminals of a three metre long wire of uniform
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- The operation of a nuclear reactor is said to be critical, if the multiplication
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the highest modulating frequency of the wave is 5 kHz, the number
- A 6 volt battery is connected to the terminals of a three metre long wire of uniform
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- The operation of a nuclear reactor is said to be critical, if the multiplication
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