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The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz incident on this metal, the cut-off voltage for the photoelectric emission is nearly
Options
(a) 2 V
(b) 3 V
(c) 5 V
(d) 1 V
Correct Answer:
2 V
Explanation:
K.E = hv – hvₜₕ = eV₀ ( V₀ = cut off voltage)
V₀ = h/e (8.2 x 10¹⁴ – 3.3 x 10¹⁴)
= 6.6 x 10⁻³⁴ x 4.9 x 10¹⁴ / 1.6 x 10⁻¹⁹ = 2 V
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Topics: Dual Nature of Matter and Radiation
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- By sucking through a straw, a student can reduce the pressure of his lungs
- The earth is assumed to be a sphere of radius R. A platform is arranged at a height R
- A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless
- An electron moving in a circular orbit of radius r makes n rotations per second.
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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