| ⇦ | 
| ⇨ |
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: - The capacitance of two concentric spherical shells of radii R₁ and R₂ (R₂>R₁) is
- For a particle in a non-uniform accelerated circular motion correct statement is
- The electric field at a distance 3R / 2 from the centre of a charged conducting
- If a steel wire of length l and magnetic moment M is bent into a semicircular arc,
- The fluctuation in the input voltage of 200 V to a domestic circuit is +10V
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The capacitance of two concentric spherical shells of radii R₁ and R₂ (R₂>R₁) is
- For a particle in a non-uniform accelerated circular motion correct statement is
- The electric field at a distance 3R / 2 from the centre of a charged conducting
- If a steel wire of length l and magnetic moment M is bent into a semicircular arc,
- The fluctuation in the input voltage of 200 V to a domestic circuit is +10V
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