Compton effect is associated with:
A.$\alpha -rays.$
B.$\beta -rays.$
C.Positive rays.
D.X-rays.
Answer
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Hint: We know that the Compton effect refers to the increase in the wavelength of photons (X-rays or gamma rays), due to their scattering by a charged particle. The impact has ended up being one of the foundations of quantum mechanics, which represents both wave and particle properties of radiation
Complete answer:
As we know, the Compton effect is used for an unusual result observed when X-rays are scattered on some materials. In Compton effect, when an electromagnetic wave is scattered off atoms, the scattered radiation is expected to be the same as the wavelength of the incident radiation. The singular protons collide with the free and loosely bonded electrons present in the atoms of the matter. Impacting photons move a portion of their energy and force to the electrons, which like this pullback. In the moment of the impact, new photons of less energy and force are delivered that dissipate at points the size of which relies upon the measure of energy lost to the withdrawing electrons.
Due to the connection between energy and frequency, the dispersed photons have a more extended frequency that additionally relies upon the size of the point through which the X-rays were redirected. The decrease in frequency, or Compton shift, doesn't rely upon the wavelength of the incident photon. The Compton dissipating is inelastic and inherent dispersing of X-rays, and it’s worth relies upon the mole part of nuclear species present in the example and the Compton Intensity of that molecule. It is the weighted entirety of Compton Intensities of individual molecules (present in the example), and the weight factor is the mole part.
Therefore, the correct answer is option D.
Note:
Remember that the Compton effect is different from photoelectric electric effect. The scattering observed in the Compton effect is inelastic. The energy associated with free electrons in case of Compton effect is mid-energy while the energy associated with electrons in photoelectric effect is low-energy.
Complete answer:
As we know, the Compton effect is used for an unusual result observed when X-rays are scattered on some materials. In Compton effect, when an electromagnetic wave is scattered off atoms, the scattered radiation is expected to be the same as the wavelength of the incident radiation. The singular protons collide with the free and loosely bonded electrons present in the atoms of the matter. Impacting photons move a portion of their energy and force to the electrons, which like this pullback. In the moment of the impact, new photons of less energy and force are delivered that dissipate at points the size of which relies upon the measure of energy lost to the withdrawing electrons.
Due to the connection between energy and frequency, the dispersed photons have a more extended frequency that additionally relies upon the size of the point through which the X-rays were redirected. The decrease in frequency, or Compton shift, doesn't rely upon the wavelength of the incident photon. The Compton dissipating is inelastic and inherent dispersing of X-rays, and it’s worth relies upon the mole part of nuclear species present in the example and the Compton Intensity of that molecule. It is the weighted entirety of Compton Intensities of individual molecules (present in the example), and the weight factor is the mole part.
Therefore, the correct answer is option D.
Note:
Remember that the Compton effect is different from photoelectric electric effect. The scattering observed in the Compton effect is inelastic. The energy associated with free electrons in case of Compton effect is mid-energy while the energy associated with electrons in photoelectric effect is low-energy.
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