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In P.E.E, Kmax of electrons is 2.0 eV. If frequency of light is decreased by 20% then Kmax is 1.0 eV. Work function of electron emitter is (in eV) :-
    Correct answer is '3'. Can you explain this answer?
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    In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreas...
    2.0 = hν - ϕ  ...... (1)
    1.0 = 0.8hν - ϕ  ...... (2)
    on solving we have, ϕ = 3 eV
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    In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreas...
    Explanation:



    Given:


    Kmax of electrons = 2.0 eV


    Frequency of light decreased by 20%.


    New Kmax is 1.0 eV



    To find: Work function of electron emitter



    Solution:


    We know that, the relation between the maximum kinetic energy of the photoelectrons and the frequency of the incident radiation is given by the Einstein’s photoelectric equation:


    Kmax = hν - Φ


    Where, h is the Planck’s constant, ν is the frequency of the incident radiation, Φ is the work function of the metal or electron emitter.


    Now, we can use the given information to find the value of work function Φ.


    According to the given information, when the frequency of the light is decreased by 20%, the new Kmax is 1.0 eV.


    Let the new frequency of light be ν'.


    Then, from the Einstein’s photoelectric equation, we have:


    1.0 eV = hν' - Φ


    We also know that the energy of a photon is given by:


    E = hν


    When the frequency of the light is decreased by 20%, the new frequency of light is:


    ν' = 0.8ν


    Substituting this value in the above equation, we get:


    1.0 eV = h(0.8ν) - Φ


    2.0 eV = hν - Φ (Given)


    Subtracting the second equation from the first equation, we get:


    1.0 eV - 2.0 eV = h(0.8ν - ν)


    -1.0 eV = -0.2hν


    hν = 5.0 eV


    Substituting this value in the second equation, we get:


    2.0 eV = 5.0 eV - Φ


    Φ = 3.0 eV



    Therefore, the work function of the electron emitter is 3.0 eV.
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    In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreased by 20% then Kmaxis 1.0 eV. Work function of electron emitter is (in eV) :-Correct answer is '3'. Can you explain this answer?
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    In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreased by 20% then Kmaxis 1.0 eV. Work function of electron emitter is (in eV) :-Correct answer is '3'. 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 In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreased by 20% then Kmaxis 1.0 eV. Work function of electron emitter is (in eV) :-Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In P.E.E, Kmaxof electrons is 2.0 eV. If frequency of light is decreased by 20% then Kmaxis 1.0 eV. Work function of electron emitter is (in eV) :-Correct answer is '3'. Can you explain this answer?.
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