Photon do not exist in reality. Newton's Corpuscular Theory of light ASSUMES an imaginary particle PHOTON to explain various phenomena of light.
I presume you asking, "How can an atom of size about 1 angstrom absorb a photon whose wavelength is 5000 angstroms? Wouldn't the photon be too large for that atom?" The paradox is resolved in this way: the instant you start to discuss electro-magnetic radiation as a photon instead of a transverse electro-magnetic wave, then you negate the wave-length aspect of the light. Instead, you view light as a collection of photons -- particles whose "size" (if that word has meaning) is point-like -- with a specific energy instead of specific wavelength. A photon is NOT a snake-like wave, vibrating like a rubber band, with a length at least that of its wave-length, as it moves through a medium. A photon is a point particle with a specific energy. You can describe light as a EM wave with a wave-length OR as a collection of point particles. You can NOT do both at the same time. Light exhibits the characteristics of one OR the other, but NEVER both.
The energy of a photon is inversely propotional to its wavelength. The wavelength of a blue photon is less than that of a red photon. That makes the blue photon more energetic. Or how about this? The energy of a photon is directly proportional to its frequency. The frequency of a blue photon is greater than that of a red photon. That makes the blue photon more energetic. The wavelength of a photon is inversely proportional to its frequency. The the longer the wavelength, the lower the frequency. The shorter the wavelength, the higher the frequency.
The opposite of a photon is an antiphoton.
Photon flux can be calculated using the formula: photon flux = v * E, where v is the frequency of the photons and E is the energy of each photon. By multiplying the frequency of the photons by the energy of each photon, you can determine the photon flux.
Photon
photon
No, the spin of different particles is a constant in each case; for example the spin of a photon is always 1.
I presume you asking, "How can an atom of size about 1 angstrom absorb a photon whose wavelength is 5000 angstroms? Wouldn't the photon be too large for that atom?" The paradox is resolved in this way: the instant you start to discuss electro-magnetic radiation as a photon instead of a transverse electro-magnetic wave, then you negate the wave-length aspect of the light. Instead, you view light as a collection of photons -- particles whose "size" (if that word has meaning) is point-like -- with a specific energy instead of specific wavelength. A photon is NOT a snake-like wave, vibrating like a rubber band, with a length at least that of its wave-length, as it moves through a medium. A photon is a point particle with a specific energy. You can describe light as a EM wave with a wave-length OR as a collection of point particles. You can NOT do both at the same time. Light exhibits the characteristics of one OR the other, but NEVER both.
A packet of light energy is called a photon.
If you know the frequency of a light wave, you can tell the wavelength, thecolor it'll appear to your eye, and the energy in each photon of the light.The energy of the wave ~APEX
The energy of a photon is inversely propotional to its wavelength. The wavelength of a blue photon is less than that of a red photon. That makes the blue photon more energetic. Or how about this? The energy of a photon is directly proportional to its frequency. The frequency of a blue photon is greater than that of a red photon. That makes the blue photon more energetic. The wavelength of a photon is inversely proportional to its frequency. The the longer the wavelength, the lower the frequency. The shorter the wavelength, the higher the frequency.
The opposite of a photon is an antiphoton.
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of the photon. Plugging in the values, the energy of a photon emitted with a wavelength of 654 nm (or 6.54 x 10^-7 m) is approximately 3.02 x 10^-19 J.
Photon flux can be calculated using the formula: photon flux = v * E, where v is the frequency of the photons and E is the energy of each photon. By multiplying the frequency of the photons by the energy of each photon, you can determine the photon flux.
tata photon plus is ratan tata and Javed Siddiqui is houner of PHoton whiz..............
. . . photon.
Photon