| ⇦ |
| ⇨ |
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: - An open knife edge of mass 200g is dropped from height 5m on a cardboard
- A block of mass 10 kg is moving in X-direction with a constant speed of 10 ms⁻¹
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- An electron having charge e and mass m is moving in a uniform electric field E.
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An open knife edge of mass 200g is dropped from height 5m on a cardboard
- A block of mass 10 kg is moving in X-direction with a constant speed of 10 ms⁻¹
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- An electron having charge e and mass m is moving in a uniform electric field E.
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