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An electron and a proton posses the same de broglie wavelength. If Ee and Ep respectively are the energy of electron and proton and v and c are their respective velocities, then Ee/Ep is equal to a)v/c b)v/2c c)v/3c d)v/4c?
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An electron and a proton posses the same de broglie wavelength. If Ee ...
(a) v/c

for kinetic energy E, Momentum P, Mp is mass of proton and Me is mass of electron ,Vp is speed of proton
for same De Broglie wavelength Lambda,

lambda =h/√(2mE)
So Ee×Me=Ep×Mp--------------(1)

also Lambda =h/mv
So
Me/Mp= Vp/Ve
Me/Mp=c/v ----------------------(2)

Solving (1)&(2) we get,
Ee/Ep=v/c
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An electron and a proton posses the same de broglie wavelength. If Ee ...
De Broglie Wavelength and Energy

The De Broglie wavelength, denoted by λ, is a concept in quantum mechanics that relates the momentum and wavelength of a particle. It is given by the equation:

λ = h / p

where λ is the De Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.

The momentum of a particle can be calculated as:

p = mv

where m is the mass of the particle and v is its velocity.

Equal De Broglie Wavelengths

Given that the electron and proton possess the same De Broglie wavelength, we can equate their momentum equations:

melectron * velectron = mproton * vproton

where melectron and mproton are the masses of the electron and proton respectively, and velectron and vproton are their velocities.

Comparing Energies

The energy of a particle can be calculated using the equation:

E = mc^2

where E is the energy, m is the mass of the particle, and c is the speed of light.

For the electron, we have:

Ee = melectron * c^2

For the proton, we have:

Ep = mproton * c^2

To compare the energies, we can calculate the ratio:

Ee / Ep = (melectron * c^2) / (mproton * c^2)

Simplifying the equation, we find:

Ee / Ep = melectron / mproton

Ratio of Electron and Proton Energies

Since the electron and proton possess the same De Broglie wavelength, their momentum equation can be rearranged as:

velectron / vproton = mproton / melectron

Substituting this into the ratio of energies equation, we get:

Ee / Ep = melectron / mproton = velectron / vproton

Therefore, the ratio of the electron energy to the proton energy is equal to the ratio of their velocities:

Ee / Ep = velectron / vproton

Answer

Therefore, the correct answer is (a) v / c. The ratio of the electron energy to the proton energy is equal to the ratio of their velocities, which is v / c.
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Read the following text and answer the following questions on the basis of the same:Electron Microscope Electron microscopes use electrons to illuminate a sample. In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera.Resolution in microscopy is limited to about half of the wavelength of the illumination source used to image the sample. Using visible light the best resolution that can be achieved by microscopes is about ~200 nm. Louis de Broglie showed that every particle or matter propagates like a wave. The wavelength of propagating electrons at a given accelerating voltage can be determined byThus, the wavelength of electrons is calculated to be 3.88 pm when the microscope is operated at 100 keV, 2. 74 pm at 200 keV and 2.24 pm at 300 keV. However, because the velocities of electrons in an electron microscope reach about 70% the speed of light with an accelerating voltage of 200 keV, there are relativistic effects on these electrons. Due to this effect, the wavelength at 100 keV, 200 keV and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively.Anyhow, the wavelength of electrons is much smaller than that of photons (2.5 pm at 200 keV). Thus if electron wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to ~0.1 nm due to the objective lens system in electron microscopes. Thus, electron microscopy can resolve subcellular structures that could not be visualized using standard fluorescence microscopy.Q. Wavelength of electron as wave at accelerating voltage 200 keV is

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An electron and a proton posses the same de broglie wavelength. If Ee and Ep respectively are the energy of electron and proton and v and c are their respective velocities, then Ee/Ep is equal to a)v/c b)v/2c c)v/3c d)v/4c?
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An electron and a proton posses the same de broglie wavelength. If Ee and Ep respectively are the energy of electron and proton and v and c are their respective velocities, then Ee/Ep is equal to a)v/c b)v/2c c)v/3c d)v/4c? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An electron and a proton posses the same de broglie wavelength. If Ee and Ep respectively are the energy of electron and proton and v and c are their respective velocities, then Ee/Ep is equal to a)v/c b)v/2c c)v/3c d)v/4c? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron and a proton posses the same de broglie wavelength. If Ee and Ep respectively are the energy of electron and proton and v and c are their respective velocities, then Ee/Ep is equal to a)v/c b)v/2c c)v/3c d)v/4c?.
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