Visible light
Neither. The beams of red light and green light will have the same number of Photons, as energy is only related to frequency. The number of Photons is dependent on the intensity of the light beams.
Photons do not come in different types like infared-photons etc. they are just the wavelength that the photons are at and nuclear fusion just happens to emit photons at a particular wavelength
Photons propagating at frequencies in the visible light spectrum can knock out electrons from atoms, known as the photoelectric effect, if their energy is greater than the photoelectric work function for that atom. However, at the energies associated with the visible light frequencies, these new photoelectrons will absorb any excess energy of the initial photons and convert it to kinetic energy, meaning that the initial photons vanish. Obviously, if the photons are gone, they can't scatter. Increasing the intensity (brightening) of the photons will cause more electrons to be emitted, but it will not increase their energy since photon energy is a function of its frequency, not quantity.Photons that retain energy after interacting with an electron via the photoelectric effect are said to undergo Compton scattering. Now, despite what everyone says, if a photon has any amount of energy greater than the applicable photoelectric work function, it can theoretically undergo Compton scattering. Yes, I'm implying that visible light can Compton scatter. However, the probability of Compton scattering at these energies is very low, not to mention these scattered photons would most likely loose all of their energy from all of the other various available atomic interactions before they could even escape the sample, which is a necessary component to measurement (something has to exist in order to be measured). Therefore, the effects of Compton scattering are negligible at visible light energies. In fact, they don't really start becoming noticeable until around energies of 100keV, which is around 105 times greater than the energies associated with visible light. These kinds of energies are associated with x-rays.
Photons are absorbed by ozone. These photons are of UV.
Photons are in action units joule-seconds.
No, microwaves are a form of electromagnetic radiationand like all electromagnetic radiation is composed of massless particles called photons or is waves depending on how it is measured.Copper is a metal.
Light is composed of quanta called photons. The more photons, the greater the intensity. To see the slightest flicker of green light (the color to which our eyes are most sensitive), the minimum number of photons is six.
The wavelength is far shorter in gamma rays, and they have a higher frequency, and their photons are much higher in energy than those of microwaves.
A hard beam is one that contains a greater number of high energy photons than low energy photons.A soft beam is one that contains a greater number of low energy photons that high energy photons.During the filtration of a heterogenous beam (one that contains photons with different energies), low energy photons are removed from the beam, effectively "hardening" it.
No. A photon has no rest mass an electron has mass.
Neither. The beams of red light and green light will have the same number of Photons, as energy is only related to frequency. The number of Photons is dependent on the intensity of the light beams.
Pretty sure it is photons.....
Longitudinal and Transverse energy waves are the two types.
Photons do not come in different types like infared-photons etc. they are just the wavelength that the photons are at and nuclear fusion just happens to emit photons at a particular wavelength
Photons propagating at frequencies in the visible light spectrum can knock out electrons from atoms, known as the photoelectric effect, if their energy is greater than the photoelectric work function for that atom. However, at the energies associated with the visible light frequencies, these new photoelectrons will absorb any excess energy of the initial photons and convert it to kinetic energy, meaning that the initial photons vanish. Obviously, if the photons are gone, they can't scatter. Increasing the intensity (brightening) of the photons will cause more electrons to be emitted, but it will not increase their energy since photon energy is a function of its frequency, not quantity.Photons that retain energy after interacting with an electron via the photoelectric effect are said to undergo Compton scattering. Now, despite what everyone says, if a photon has any amount of energy greater than the applicable photoelectric work function, it can theoretically undergo Compton scattering. Yes, I'm implying that visible light can Compton scatter. However, the probability of Compton scattering at these energies is very low, not to mention these scattered photons would most likely loose all of their energy from all of the other various available atomic interactions before they could even escape the sample, which is a necessary component to measurement (something has to exist in order to be measured). Therefore, the effects of Compton scattering are negligible at visible light energies. In fact, they don't really start becoming noticeable until around energies of 100keV, which is around 105 times greater than the energies associated with visible light. These kinds of energies are associated with x-rays.
Photons are absorbed by ozone. These photons are of UV.
Photons are pieces of light. If you see a light, then there are photons.