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When a 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)4 eV
- b)2 eV
- c)1.5 eV
- d)3 eV
Correct answer is option 'A'. Can you explain this answer?
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When a photon of energy 4.0 eV strikes the surface of a metal A, the ...
De-Broglie wavelength (λ),
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Momentum,
On solving we get, TA = 2 eV
∴ KEB = TA – 1.5 = 2 – 1.5 = 0.5 eV
∴ Work function of metal B is
φB = EB – KEB = 4.5 – 0.5 = 4 eV
Most Upvoted Answer
When a photon of energy 4.0 eV strikes the surface of a metal A, the ...
Given:
Photon energy of incident light on metal A = 4.0 eV
Maximum kinetic energy of photoelectrons from metal A = TA eV
De-Broglie wavelength of photoelectrons from metal A = λA
Maximum kinetic energy of photoelectrons from metal B = TB = (TA – 1.5) eV
De-Broglie wavelength of photoelectrons from metal B = λB = 2λA
To find:
The work function of metal B
Solution:
1. Relation between photon energy and kinetic energy of photoelectrons:
The energy of a photon (E) can be related to the kinetic energy (K) of a photoelectron using the equation:
E = K + Φ
where Φ is the work function of the metal.
2. Relation between de-Broglie wavelength and kinetic energy of photoelectrons:
The de-Broglie wavelength (λ) of a particle can be related to its kinetic energy (K) using the equation:
λ = h / √(2mK)
where h is the Planck's constant and m is the mass of the particle.
3. Applying the above relations:
For metal A:
E = 4.0 eV
K = TA eV
λ = λA
For metal B:
E = 4.50 eV
K = TB = (TA – 1.5) eV
λ = λB = 2λA
4. Using the relation E = K + Φ:
For metal A:
4.0 eV = TA eV + ΦA
For metal B:
4.50 eV = (TA – 1.5) eV + ΦB
5. Using the relation λ = h / √(2mK):
For metal A:
λA = h / √(2mA TA)
For metal B:
2λA = h / √(2mB (TA – 1.5))
6. Comparing the two equations:
√(2mB (TA – 1.5)) = 2√(2mA TA)
Squaring both sides:
2mB (TA – 1.5) = 4(2mA TA)
mB (TA – 1.5) = 2mA TA
mB TA – 1.5mB = 2mA TA
mB TA – 2mA TA = 1.5mB
TA (mB – 2mA) = 1.5mB
TA = (1.5mB) / (mB – 2mA)
7. Substituting the value of TA in the equation E = K + Φ:
For metal A:
4.0 eV = TA eV + ΦA
ΦA = 4.0 eV – TA eV
For metal B:
4.50 eV = (TA – 1.5) eV + ΦB
ΦB = 4.50 eV – (TA – 1.5) eV
Φ
Photon energy of incident light on metal A = 4.0 eV
Maximum kinetic energy of photoelectrons from metal A = TA eV
De-Broglie wavelength of photoelectrons from metal A = λA
Maximum kinetic energy of photoelectrons from metal B = TB = (TA – 1.5) eV
De-Broglie wavelength of photoelectrons from metal B = λB = 2λA
To find:
The work function of metal B
Solution:
1. Relation between photon energy and kinetic energy of photoelectrons:
The energy of a photon (E) can be related to the kinetic energy (K) of a photoelectron using the equation:
E = K + Φ
where Φ is the work function of the metal.
2. Relation between de-Broglie wavelength and kinetic energy of photoelectrons:
The de-Broglie wavelength (λ) of a particle can be related to its kinetic energy (K) using the equation:
λ = h / √(2mK)
where h is the Planck's constant and m is the mass of the particle.
3. Applying the above relations:
For metal A:
E = 4.0 eV
K = TA eV
λ = λA
For metal B:
E = 4.50 eV
K = TB = (TA – 1.5) eV
λ = λB = 2λA
4. Using the relation E = K + Φ:
For metal A:
4.0 eV = TA eV + ΦA
For metal B:
4.50 eV = (TA – 1.5) eV + ΦB
5. Using the relation λ = h / √(2mK):
For metal A:
λA = h / √(2mA TA)
For metal B:
2λA = h / √(2mB (TA – 1.5))
6. Comparing the two equations:
√(2mB (TA – 1.5)) = 2√(2mA TA)
Squaring both sides:
2mB (TA – 1.5) = 4(2mA TA)
mB (TA – 1.5) = 2mA TA
mB TA – 1.5mB = 2mA TA
mB TA – 2mA TA = 1.5mB
TA (mB – 2mA) = 1.5mB
TA = (1.5mB) / (mB – 2mA)
7. Substituting the value of TA in the equation E = K + Φ:
For metal A:
4.0 eV = TA eV + ΦA
ΦA = 4.0 eV – TA eV
For metal B:
4.50 eV = (TA – 1.5) eV + ΦB
ΦB = 4.50 eV – (TA – 1.5) eV
Φ
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When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer?
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When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer?.
When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer?.
Solutions for When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice When a 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)4 eVb)2 eVc)1.5 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice NEET tests.
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