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When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer?.

Solutions for When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer?, a detailed solution for When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? has been provided alongside types of When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _________(Round off to the nearest integer)[use √3= 1.73, h = 6.63 x 10-34 Js, me = 9.1 x 10-31 kg; c = 3.0 x 108 ms-1; 1eV = 1.6 x 10-19J]Correct answer is '9'. Can you explain this answer? tests, examples and also practice JEE tests.