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When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]
- a)2K
- b)K
- c)K + hv
- d)K + E0
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
When photons of energy hv fall on an aluminiumplate (of work function ...
Applying Einstein's formula for photoelectricity
If we use 2v frequency then let the kinetic
energy becomes K'
So,
energy becomes K'
So,
Most Upvoted Answer
When photons of energy hv fall on an aluminiumplate (of work function ...
Explanation:
The energy of a photon is given by E = hv, where h is Planck's constant and v is the frequency of the radiation. When a photon falls on a metal plate, it can eject photoelectrons if its energy is greater than or equal to the work function of the metal (E0). The maximum kinetic energy of the ejected photoelectrons is given by the equation:
Kmax = E - E0
where Kmax is the maximum kinetic energy of the ejected photoelectrons, E is the energy of the incident photon, and E0 is the work function of the metal.
Now, let's consider what happens when the frequency of the radiation is doubled. According to the equation E = hv, doubling the frequency means doubling the energy of each photon. Therefore, the energy of the incident photon is now 2hv. Using the equation for the maximum kinetic energy of the ejected photoelectrons, we get:
Kmax' = 2hv - E0
Simplifying this equation, we get:
Kmax' = 2(E - E0)
Substituting the equation for E, we get:
Kmax' = 2Kmax
Therefore, when the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons is doubled as well. The correct answer is option C, K hv.
The energy of a photon is given by E = hv, where h is Planck's constant and v is the frequency of the radiation. When a photon falls on a metal plate, it can eject photoelectrons if its energy is greater than or equal to the work function of the metal (E0). The maximum kinetic energy of the ejected photoelectrons is given by the equation:
Kmax = E - E0
where Kmax is the maximum kinetic energy of the ejected photoelectrons, E is the energy of the incident photon, and E0 is the work function of the metal.
Now, let's consider what happens when the frequency of the radiation is doubled. According to the equation E = hv, doubling the frequency means doubling the energy of each photon. Therefore, the energy of the incident photon is now 2hv. Using the equation for the maximum kinetic energy of the ejected photoelectrons, we get:
Kmax' = 2hv - E0
Simplifying this equation, we get:
Kmax' = 2(E - E0)
Substituting the equation for E, we get:
Kmax' = 2Kmax
Therefore, when the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons is doubled as well. The correct answer is option C, K hv.
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When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer?.
When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer?.
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When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer?, a detailed solution for When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of When photons of energy hv fall on an aluminiumplate (of work function E0), photoelectrons ofmaximum kinetic energy K are ejected. If thefrequency of the radiation is doubled, themaximum kinetic energy of the ejectedphotoelectrons will be [2006]a)2Kb)Kc)K + hvd)K + E0Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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