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When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. 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 photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. 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 photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer?.

Solutions for When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. 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 photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, thena)The work function of A is 2.25 eVb)The work function of B is 4.20 eVc)TA = 2.00 eVd)TB = 2.75 eVCorrect answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice JEE tests.