| ⇦ |
| ⇨ |
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 cricket ball of mass 250 g collides with a bat with velocity 10 m/s
- The current flowing through a lamp marked as 50 W and 250 V is
- A step down transformer is used on a 1000 V line to deliver 20 A at 120 V
- A change of 0.04 V takes place between the base and the emitter when an input
- A capacitor is charged to 200 volt. It has a charge of 0.1 coulomb.
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A cricket ball of mass 250 g collides with a bat with velocity 10 m/s
- The current flowing through a lamp marked as 50 W and 250 V is
- A step down transformer is used on a 1000 V line to deliver 20 A at 120 V
- A change of 0.04 V takes place between the base and the emitter when an input
- A capacitor is charged to 200 volt. It has a charge of 0.1 coulomb.
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