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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.
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Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When alpha particles captures and electron, then it becomes a
- The de-Broglie wavelength of a proton(charge = 1.6 x 10⁻¹⁹ C, Mass = 1.6 x 10⁻²⁷ kg)
- The radius of a nucleus of mass number A is directly proportional to
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- The speed of light in media M₁ and M₂ are 1.5×10⁸ ms⁻¹ and 2×10⁸ ms⁻¹ respectively.
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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