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If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ , then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :
- a)2/3 λ
- b)4/9 λ
- c)3/4 λ
- d)1/2 λ
Correct answer is option 'B'. Can you explain this answer?
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If p is the momentum of the fastest electron ejected from a metal surf...
W0 is too small in comparision to K.E.
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If p is the momentum of the fastest electron ejected from a metal surf...
If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, we can use the equation for the momentum of a particle, p = mv, where m is the mass and v is the velocity.
The wavelength of light, λ, can be related to the velocity of the electron using the de Broglie wavelength equation, λ = h/p, where h is the Planck's constant.
Combining these two equations, we can solve for the velocity of the electron. Rearranging the de Broglie equation, we have p = h/λ, which can be rearranged as pλ = h.
Since the momentum, p, is given as the momentum of the fastest electron, we can substitute p with its value in the equation above, giving us (pf)λ = h, where pf is the momentum of the fastest electron.
Simplifying, we get pf = h/λ.
Therefore, the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ is given by pf = h/λ.
The wavelength of light, λ, can be related to the velocity of the electron using the de Broglie wavelength equation, λ = h/p, where h is the Planck's constant.
Combining these two equations, we can solve for the velocity of the electron. Rearranging the de Broglie equation, we have p = h/λ, which can be rearranged as pλ = h.
Since the momentum, p, is given as the momentum of the fastest electron, we can substitute p with its value in the equation above, giving us (pf)λ = h, where pf is the momentum of the fastest electron.
Simplifying, we get pf = h/λ.
Therefore, the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ is given by pf = h/λ.
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If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :a)2/3λb)4/9λc)3/4λd)1/2λCorrect answer is option 'B'. Can you explain this answer?
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If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :a)2/3λb)4/9λc)3/4λd)1/2λCorrect 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 If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :a)2/3λb)4/9λc)3/4λd)1/2λCorrect 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 If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :a)2/3λb)4/9λc)3/4λd)1/2λCorrect answer is option 'B'. Can you explain this answer?.
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