No, not all photons released can be seen as visible light. Photons can have a range of energies and wavelengths, from radio waves at one end of the spectrum to gamma rays at the other. Visible light falls within a specific range of wavelengths that are detectable by the human eye, but there are many other types of photons that are not visible to us.
Within the range of visible light, it's the light with the shortest wavelength (highest frequency). That's theviolet end of the spectrum. Past that, ultraviolet light, which is invisible, has higher energy.
When an electron falls from a higher energy level to a lower energy level, the energy it was carrying is released in the form of a photon. The energy of the photon is equal to the difference in energy between the two levels. This released energy can be observed as light emission in the visible or invisible spectra, depending on the specific energy levels involved.
Infrared radiation has less energy (per photon) than visible light.
The wavelength λ of a photon can be calculated using the energy of the photon E and the speed of light c, where λ = c/E. The energy of the photon depends on the emission process that released it.
A microwave signal at 50 GHz has waves that are 10,000 times as long as a visible signal at yellow (600 nm) has. Therefore the yellow photon carries 10,000 times as much energy as the 50 GHz photon does.
UV has higher energy (per photon) than visible light.
Within the range of visible light, it's the light with the shortest wavelength (highest frequency). That's theviolet end of the spectrum. Past that, ultraviolet light, which is invisible, has higher energy.
A photon's energy is directly proportional to its frequency (inversely proportional to its wavelength).In any given interval of the spectrum, the highest frequency (shortest wavelength) carries the most energy.For visible light, that corresponds to the violet end of the 'rainbow'. The last color your eyes can perceiveat that end is the color with the most energy per photon.
Ultraviolet light has a greater energy per photon.
Infrared light has lower energy compared to visible light. This is because infrared light has longer wavelengths, which correspond to lower frequencies and energies.
When an electron falls from a higher energy level to a lower energy level, the energy it was carrying is released in the form of a photon. The energy of the photon is equal to the difference in energy between the two levels. This released energy can be observed as light emission in the visible or invisible spectra, depending on the specific energy levels involved.
It is not meaningful to talk about "amplitude of the visible light spectrum". One might think that more intense light would mean greater amplitude of the light wave, but it just means more photons. "Visible light" is made up of photons. A single photon has a certain quantifiable energy, and that energy is discussed in terms of frequency or wavelength. A photon with low frequency (towards the red end of the visible light spectrum, for instance) is less energetic than a photon with high frequency (towards the blue end and beyond). For all intents and purposes, the amplitude of a photon wave-packet could be said to be of "unit amplitude", the amplitude of light.
Infrared radiation has less energy (per photon) than visible light.
Well, first of all, protons don't make light. I think you mean 'photons'. A photon of ultraviolet light carries more energy than a photon of visible light, because it has a higher frequency / shorter wavelength.
Visible light has a higher frequency, a higher energy per photon, and a smaller wavelength, compared to infrared.
no, sound and light are very different Sound is a vibration in air, light is a particle called a photon moving through the air
light is given off by an atom when and electron moves from one shell to a lower shell and a specific amount of energy is released in the process (known as a photon). If the wavelength of the released photon are in the spectrum of visible light, we will see it as a specific color based on the wavelength of the photon.