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When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is
- a)1.5 eV
- b)4 eV
- c)3 eV
- d)2 eV
Correct answer is option 'B'. Can you explain this answer?
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When photon of energy 4.0 eV strikes the surface of a metal A, the eje...
The kinetic energy of the ejected photoelectrons can be calculated using the equation:
KE = E - Φ
where KE is the kinetic energy, E is the energy of the incident photon, and Φ is the work function of the metal. In this case, the energy of the incident photon is 4.0 eV.
Now, the de Broglie wavelength of the ejected photoelectrons can be calculated using the equation:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the electron. The momentum of the electron can be calculated using the equation:
p = √(2mKE)
where m is the mass of the electron (9.109 x 10^-31 kg).
Given that the maximum kinetic energy of the ejected photoelectrons is TA eV, we can substitute this value into the equation to calculate the momentum, and then substitute the momentum into the de Broglie wavelength equation to find the wavelength.
KE = E - Φ
where KE is the kinetic energy, E is the energy of the incident photon, and Φ is the work function of the metal. In this case, the energy of the incident photon is 4.0 eV.
Now, the de Broglie wavelength of the ejected photoelectrons can be calculated using the equation:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the electron. The momentum of the electron can be calculated using the equation:
p = √(2mKE)
where m is the mass of the electron (9.109 x 10^-31 kg).
Given that the maximum kinetic energy of the ejected photoelectrons is TA eV, we can substitute this value into the equation to calculate the momentum, and then substitute the momentum into the de Broglie wavelength equation to find the wavelength.
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When photon of energy 4.0 eV strikes the surface of a metal A, the eje...
de Broglie wavelength (λ),
On solving, TA = 2 eV
∴ KB = TA – 1.5 = 0.5 eV
∴ Work function of metal B is:
ϕB = EB – KB
= 4.5 – 0.5 = 4 eV
For A, ϕA = EA – TA = 2 eV
So, work function for B is 4 eV
On solving, TA = 2 eV
∴ KB = TA – 1.5 = 0.5 eV
∴ Work function of metal B is:
ϕB = EB – KB
= 4.5 – 0.5 = 4 eV
For A, ϕA = EA – TA = 2 eV
So, work function for B is 4 eV
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When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer?
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When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. 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 photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. 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 photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer?.
When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. 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 photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. 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 photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer?.
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When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B isa)1.5 eVb)4 eVc)3 eVd)2 eVCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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