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A light of wavelength 5000Å falls on a sensitive plate with photoelectric work function 1.90 eV. Kinetic energy of the emitted photoelectrons will be (given, h=6.62×10⁻³⁴ Js)
Options
(a) 0.1 eV
(b) 2 eV
(c) 0.58 eV
(d) 1.581 eV
Correct Answer:
0.58 eV
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
- To demonstrate the phenomenon of interference, we require two sources which emit
- If the minimum energy of photons needed to produce photoelectrons is 3 eV,
- Curie is the unit of
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
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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