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Can you explain the answer of this question below:
Photon and electron are given same energy (10-20 J) Wavelength associated with photon statement will he and electron are λp and λe the correct
  • A:
    λp > λe
  • B:
    λp < λe
  • C:
    λp = λe
  • D:
    λe / λp = c
The answer is a.
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Most Upvoted Answer
Can you explain the answer of this question below:Photon and electron ...
Since Both are provided with same energy and it's the large amount of energy (as per the question given 10-20J) given to the microscopic particles, here kinetic energy dominates. Hence the debroglie wavelength inversely proportional to mass and photons are mass less, so photon has longer wavelength as electron is massive than photon. Hope you get it!
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Can you explain the answer of this question below:Photon and electron ...
The question is asking about the wavelength associated with a photon and an electron that both have the same energy of 10-20 Joules (J).

First, let's understand the concept of energy and wavelength in the context of photons and electrons:

- Photons are particles of light and have both particle-like and wave-like properties. The energy of a photon is directly related to its wavelength, according to the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. This equation shows that as the energy of a photon increases, its wavelength decreases.

- Electrons, on the other hand, are subatomic particles that have both particle-like and wave-like properties. However, their energy is not directly related to their wavelength like photons. Instead, the energy of an electron is determined by its kinetic energy, which depends on its velocity.

Given that the photon and electron have the same energy of 10-20 J, we cannot directly determine their respective wavelengths without additional information. The energy of the photon can be used to calculate its wavelength using the equation mentioned above. However, the energy of the electron does not provide any information about its wavelength.

In conclusion, the wavelength associated with the photon can be determined using its energy, but the wavelength associated with the electron cannot be determined based solely on its energy.
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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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Can you explain the answer of this question below:Photon and electron are given same energy (10-20J) Wavelength associated with photon statement will he and electron are λpand λethe correctA:λp> λeB:λp< λeC:λp= λeD:λe/ λp= cThe answer is a.
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