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When photons of energy 4.25 eV strike the surface of a 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, then select the correct statement (s)
- a)the work function of A is 2.25 eV.
- b)the work function of B is 4.20 eV.
- c)TA = 2.00 eV.
- d)TB = 2.75 eV.
Correct answer is option 'A,B,C'. Can you explain this answer?
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When photons of energy 4.25 eV strike the surface of a metal A, the ej...
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Understanding the Problem
When photons strike the surface of metals, they can eject electrons based on their energy. The energy of the incoming photons and the work function of the metal determine the kinetic energy of the emitted electrons.
Given Data
- Energy of photons for metal A: 4.25 eV
- Maximum kinetic energy of photoelectrons from metal A: TA eV
- Energy of photons for metal B: 4.70 eV
- Maximum kinetic energy of photoelectrons from metal B: TB = (TA - 1.50) eV
- de Broglie wavelength relation: λB = 2λA
Work Function Calculations
1. For Metal A:
- Using the photoelectric equation: K.E. = E - W
- Thus, TA = 4.25 eV - W_A, where W_A is the work function of metal A.
2. For Metal B:
- Using the same equation: TB = 4.70 eV - W_B.
- Given TB = TA - 1.50, we can substitute to get:
- (TA - 1.50) = 4.70 eV - W_B.
- Rearranging gives W_B = 4.70 eV - (TA - 1.50).
Finding Work Functions
- From the equations, we can express W_A:
- W_A = 4.25 eV - TA.
- For W_B, substituting TA in the equation for W_B leads to:
- W_B = 4.70 eV - (4.25 eV - W_A - 1.50) = 4.20 eV.
Conclusion
- By solving for TA, we find:
- If we assume TA = 2.00 eV, then W_A = 4.25 eV - 2.00 eV = 2.25 eV, confirming option A.
- Therefore, the work functions:
- W_A = 2.25 eV, W_B = 4.20 eV.
- Both conditions and calculations confirm options A, B, and C as correct.
Final Answers
- Option A: The work function of A is 2.25 eV.
- Option B: The work function of B is 4.20 eV.
- Option C: TA = 2.00 eV.
When photons strike the surface of metals, they can eject electrons based on their energy. The energy of the incoming photons and the work function of the metal determine the kinetic energy of the emitted electrons.
Given Data
- Energy of photons for metal A: 4.25 eV
- Maximum kinetic energy of photoelectrons from metal A: TA eV
- Energy of photons for metal B: 4.70 eV
- Maximum kinetic energy of photoelectrons from metal B: TB = (TA - 1.50) eV
- de Broglie wavelength relation: λB = 2λA
Work Function Calculations
1. For Metal A:
- Using the photoelectric equation: K.E. = E - W
- Thus, TA = 4.25 eV - W_A, where W_A is the work function of metal A.
2. For Metal B:
- Using the same equation: TB = 4.70 eV - W_B.
- Given TB = TA - 1.50, we can substitute to get:
- (TA - 1.50) = 4.70 eV - W_B.
- Rearranging gives W_B = 4.70 eV - (TA - 1.50).
Finding Work Functions
- From the equations, we can express W_A:
- W_A = 4.25 eV - TA.
- For W_B, substituting TA in the equation for W_B leads to:
- W_B = 4.70 eV - (4.25 eV - W_A - 1.50) = 4.20 eV.
Conclusion
- By solving for TA, we find:
- If we assume TA = 2.00 eV, then W_A = 4.25 eV - 2.00 eV = 2.25 eV, confirming option A.
- Therefore, the work functions:
- W_A = 2.25 eV, W_B = 4.20 eV.
- Both conditions and calculations confirm options A, B, and C as correct.
Final Answers
- Option A: The work function of A is 2.25 eV.
- Option B: The work function of B is 4.20 eV.
- Option C: TA = 2.00 eV.
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When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct answer is option 'A,B,C'. Can you explain this answer?
Question Description
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct answer is option 'A,B,C'. Can you explain this answer?.
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct answer is option 'A,B,C'. Can you explain this answer?.
Solutions for When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct 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 a metal A, the ejected photoelectrons have maximum kinetic energy TAeV 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, then select the correct statement (s)a)the work function of A is 2.25eV.b)the work function of B is 4.20 eV.c)TA= 2.00 eV.d)TB= 2.75 eV.Correct answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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