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Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.
The difference in maximum kinetic energy of photoelectrons from A and from B
- a)Increases with increase in E
- b)Decreases with increase in E
- c)First increases than decrease with increase in E.
- d)Remains constant with increase in E
Correct answer is option 'D'. Can you explain this answer?
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Work function of metal A is equal to the ionization energy of hydroge...
(13.6) − (3.2) = 10.2 eV
= constant
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Work function of metal A is equal to the ionization energy of hydroge...
Explanation:
- The work function of a metal is the minimum energy required to remove an electron from its surface.
- The ionization energy of an atom or ion is the energy required to remove an electron from it.
- In this question, the work function of metal A is equal to the ionization energy of a hydrogen atom in its first excited state. The ionization energy of hydrogen in its first excited state is equal to the energy difference between the first excited state and the ground state.
- Similarly, the work function of metal B is equal to the ionization energy of a helium ion in its second orbit. The ionization energy of a helium ion in its second orbit is equal to the energy difference between the second orbit and the first orbit.
- Photons of the same energy E are incident on both metal A and metal B.
- The maximum kinetic energy of the photoelectrons emitted from metal A is twice that of the photoelectrons emitted from metal B.
- The maximum kinetic energy of a photoelectron can be calculated using the equation: KEmax = hν - φ, where KEmax is the maximum kinetic energy, h is Planck's constant, ν is the frequency of the incident photons, and φ is the work function.
- Since the energy of the incident photons is the same for both metals, E = hν, the equation for the maximum kinetic energy becomes: KEmax = E - φ.
- Given that KEmax(A) = 2 * KEmax(B), we can write the equation as: E - φ(A) = 2 * (E - φ(B)).
- Rearranging the equation, we get: φ(A) - φ(B) = E.
- This equation shows that the difference in the work functions of metal A and metal B is equal to the energy of the incident photons.
- Since the energy of the incident photons is constant, the difference in the maximum kinetic energy of the photoelectrons from metal A and metal B also remains constant with an increase in E.
- Therefore, the correct answer is option 'D' - the difference in maximum kinetic energy of photoelectrons from A and B remains constant with an increase in E.
- The work function of a metal is the minimum energy required to remove an electron from its surface.
- The ionization energy of an atom or ion is the energy required to remove an electron from it.
- In this question, the work function of metal A is equal to the ionization energy of a hydrogen atom in its first excited state. The ionization energy of hydrogen in its first excited state is equal to the energy difference between the first excited state and the ground state.
- Similarly, the work function of metal B is equal to the ionization energy of a helium ion in its second orbit. The ionization energy of a helium ion in its second orbit is equal to the energy difference between the second orbit and the first orbit.
- Photons of the same energy E are incident on both metal A and metal B.
- The maximum kinetic energy of the photoelectrons emitted from metal A is twice that of the photoelectrons emitted from metal B.
- The maximum kinetic energy of a photoelectron can be calculated using the equation: KEmax = hν - φ, where KEmax is the maximum kinetic energy, h is Planck's constant, ν is the frequency of the incident photons, and φ is the work function.
- Since the energy of the incident photons is the same for both metals, E = hν, the equation for the maximum kinetic energy becomes: KEmax = E - φ.
- Given that KEmax(A) = 2 * KEmax(B), we can write the equation as: E - φ(A) = 2 * (E - φ(B)).
- Rearranging the equation, we get: φ(A) - φ(B) = E.
- This equation shows that the difference in the work functions of metal A and metal B is equal to the energy of the incident photons.
- Since the energy of the incident photons is constant, the difference in the maximum kinetic energy of the photoelectrons from metal A and metal B also remains constant with an increase in E.
- Therefore, the correct answer is option 'D' - the difference in maximum kinetic energy of photoelectrons from A and B remains constant with an increase in E.
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Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer?
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Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. 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 Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer?.
Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. 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 Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer?.
Solutions for Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.The difference in maximum kinetic energy of photoelectrons from A and from Ba)Increases with increase in Eb)Decreases with increase in Ec)First increases than decrease with increase in E.d)Remains constant with increase in ECorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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