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When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λA then which is not correct?
- a)The work function of A is 2.25 eV
- b)The work function of B is 3.70 eV
- c)TA = 2.00eV
- d)TB = 0.5eV
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
When photons of energy 4.25eV strike the surface of a metal A, the eje...
Let work function of A and B be wA and wB and TA, TB are kinetic energy
∴ 4.25 = wA + TA
or TA=4.25-wA
similarly TB=4.70-wB
∴ TB-TA=0.45+wA-wB
-1.5=0.45+wA-wB (∴TB-TA=-1.5)
or wB-wA= 1.95
∴
∴
(K is kinetic energy)
∴
Also
∴
∴ TA= 2eV TB= 0.5eV wA = 2.25 eV wB = 4.2 eV
∴ 4.25 = wA + TA
or TA=4.25-wA
similarly TB=4.70-wB
∴ TB-TA=0.45+wA-wB
-1.5=0.45+wA-wB (∴TB-TA=-1.5)
or wB-wA= 1.95
∴
∴
(K is kinetic energy)∴
Also
∴
∴ TA= 2eV TB= 0.5eV wA = 2.25 eV wB = 4.2 eV
Most Upvoted Answer
When photons of energy 4.25eV strike the surface of a metal A, the eje...
To find the maximum kinetic energy of the ejected photoelectrons, we can use the equation:
KE = hf - Φ
Where KE is the kinetic energy of the ejected electron, hf is the energy of the incident photon (4.25 eV in this case), and Φ is the work function of the metal A.
The de Broglie wavelength of the ejected photoelectron 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.
Since the momentum of an electron is given by p = √(2mE), where m is the mass of the electron (9.11 x 10^-31 kg) and E is the kinetic energy of the electron, we can substitute the value of E into the equation to get:
p = √(2mKE)
Now, let's calculate the maximum kinetic energy and de Broglie wavelength:
KE = hf - Φ = (4.25 eV) - Φ
To convert eV to joules, we use the conversion factor: 1 eV = 1.602 x 10^-19 J
KE = (4.25 eV) * (1.602 x 10^-19 J/eV) - Φ
Now, let's calculate the de Broglie wavelength:
p = √(2mKE) = √(2 * (9.11 x 10^-31 kg) * KE)
Finally, we can calculate the de Broglie wavelength using the equation:
λ = h / p = (6.626 x 10^-34 J·s) / p
Remember to substitute the value of p from the previous calculation.
Please note that we need the value of the work function (Φ) for metal A in order to provide a complete answer.
KE = hf - Φ
Where KE is the kinetic energy of the ejected electron, hf is the energy of the incident photon (4.25 eV in this case), and Φ is the work function of the metal A.
The de Broglie wavelength of the ejected photoelectron 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.
Since the momentum of an electron is given by p = √(2mE), where m is the mass of the electron (9.11 x 10^-31 kg) and E is the kinetic energy of the electron, we can substitute the value of E into the equation to get:
p = √(2mKE)
Now, let's calculate the maximum kinetic energy and de Broglie wavelength:
KE = hf - Φ = (4.25 eV) - Φ
To convert eV to joules, we use the conversion factor: 1 eV = 1.602 x 10^-19 J
KE = (4.25 eV) * (1.602 x 10^-19 J/eV) - Φ
Now, let's calculate the de Broglie wavelength:
p = √(2mKE) = √(2 * (9.11 x 10^-31 kg) * KE)
Finally, we can calculate the de Broglie wavelength using the equation:
λ = h / p = (6.626 x 10^-34 J·s) / p
Remember to substitute the value of p from the previous calculation.
Please note that we need the value of the work function (Φ) for metal A in order to provide a complete answer.
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When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer?
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When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 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 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer?.
When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 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 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer?.
Solutions for When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Here you can find the meaning of When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de broglie wavelength,λA, The maximum kinetic energy of photoelectrons liberated from anothermetal B by photons of energy 4.70V is TB = TA - 1.50 eV. If the de Broglie wavelength of these photoelectrons is =,λB = 2λAthen which is not correct? a)The work function of A is 2.25 eV b)The work function of B is 3.70 eV c)TA = 2.00eV d)TB = 0.5eVCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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