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Linear momenta of a proton and an electron are equal. Relative to an electron
- a)de-Broglie wavelength of proton is more
- b)Kinetic energy of proton is more
- c)de-Broglie wavelength of proton and electron are equal
- d)de-Broglie wavelength of proton is less
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Linear momenta of a proton and an electron are equal. Relative to an e...
If pe = pp,
therefore, de-Broglie wavelength of proton and electron are equal.
The correct answer is: de-Broglie wavelength of proton and electron are equal
The correct answer is: de-Broglie wavelength of proton and electron are equal
Most Upvoted Answer
Linear momenta of a proton and an electron are equal. Relative to an e...
Relative to an electron, the linear momenta of a proton and an electron are equal. This means that the magnitude of the momentum of the proton is the same as that of the electron, but their directions are opposite.
Explanation:
The linear momentum (p) of a particle is given by the product of its mass (m) and velocity (v): p = mv.
Given that the linear momenta of the proton and electron are equal, we can write:
mpvp = meve,
where mp and me are the masses of the proton and electron respectively, and vp and ve are their velocities.
Now, let's consider the de Broglie wavelength for the proton and electron.
The de Broglie wavelength (λ) of a particle is given by the ratio of its Planck's constant (h) to its momentum: λ = h/p.
For the proton, the de Broglie wavelength is given by λp = h/mpvp.
For the electron, the de Broglie wavelength is given by λe = h/meve.
Since mpvp = meve, we can see that the de Broglie wavelengths of the proton and electron are equal: λp = λe.
Therefore, the correct answer is option C: the de Broglie wavelength of the proton and electron are equal.
This result is consistent with the wave-particle duality of quantum mechanics, which states that particles can exhibit both wave-like and particle-like properties. The de Broglie wavelength is a measure of the wave-like nature of a particle, and in this case, it is the same for both the proton and electron.
Explanation:
The linear momentum (p) of a particle is given by the product of its mass (m) and velocity (v): p = mv.
Given that the linear momenta of the proton and electron are equal, we can write:
mpvp = meve,
where mp and me are the masses of the proton and electron respectively, and vp and ve are their velocities.
Now, let's consider the de Broglie wavelength for the proton and electron.
The de Broglie wavelength (λ) of a particle is given by the ratio of its Planck's constant (h) to its momentum: λ = h/p.
For the proton, the de Broglie wavelength is given by λp = h/mpvp.
For the electron, the de Broglie wavelength is given by λe = h/meve.
Since mpvp = meve, we can see that the de Broglie wavelengths of the proton and electron are equal: λp = λe.
Therefore, the correct answer is option C: the de Broglie wavelength of the proton and electron are equal.
This result is consistent with the wave-particle duality of quantum mechanics, which states that particles can exhibit both wave-like and particle-like properties. The de Broglie wavelength is a measure of the wave-like nature of a particle, and in this case, it is the same for both the proton and electron.
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Linear momenta of a proton and an electron are equal. Relative to an e...
If pe = pp,
therefore, de-Broglie wavelength of proton and electron are equal.
The correct answer is: de-Broglie wavelength of proton and electron are equal
The correct answer is: de-Broglie wavelength of proton and electron are equal
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Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer?
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Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer?.
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Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Linear momenta of a proton and an electron are equal. Relative to an electrona)de-Broglie wavelength of proton is moreb)Kinetic energy of proton is morec)de-Broglie wavelength of proton and electron are equald)de-Broglie wavelength of proton is lessCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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