| ⇦ | 
| ⇨ |
Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹. Assuming the specific charge of the electron to be 1.8×10¹¹ C kg⁻¹, the value of the stopping potential in volt will be
Options
(a) 2
(b) 3
(c) 4
(d) 6
Correct Answer:
4
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A proton beam enters a magnetic field of 10⁻⁴ Wb/m² normally. If the specific charge
- The oscillation of a body on a smooth horizontal surface is represented by the equation
- A spherical shell of radius 10cm is carrying a charge q. If the electric potential
- Thin uniform rod of length L, cross-sectional area A and density ρ is rotated
- When a wire of uniform cross-section a, length l and resistance R
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A proton beam enters a magnetic field of 10⁻⁴ Wb/m² normally. If the specific charge
- The oscillation of a body on a smooth horizontal surface is represented by the equation
- A spherical shell of radius 10cm is carrying a charge q. If the electric potential
- Thin uniform rod of length L, cross-sectional area A and density ρ is rotated
- When a wire of uniform cross-section a, length l and resistance R
Topics: Dual Nature of Matter and Radiation (150)
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
ev=(1/2)mV^2max
v= (mV^2max)/2e
=(V^2max)/2 (e/m)
=((1.2×10^6)^2)/(2×1.8×10^11)=4v