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One milliwatt of light of wavelength 4560 is incident on a cesium surface. Calculate the photoelectric current (in microampere) liberated assuming a quantum efficiency of 0.5%, given Planck’s constant h = 6.62 * 1O-34 J-s and velocity of light c = 3 * 108 m/sec.Correct answer is '1.84'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about One milliwatt of light of wavelength 4560 is incident on a cesium surface. Calculate the photoelectric current (in microampere) liberated assuming a quantum efficiency of 0.5%, given Planck’s constant h = 6.62 * 1O-34 J-s and velocity of light c = 3 * 108 m/sec.Correct answer is '1.84'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One milliwatt of light of wavelength 4560 is incident on a cesium surface. Calculate the photoelectric current (in microampere) liberated assuming a quantum efficiency of 0.5%, given Planck’s constant h = 6.62 * 1O-34 J-s and velocity of light c = 3 * 108 m/sec.Correct answer is '1.84'. Can you explain this answer?.

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