Photon Kinetic Energy: Wavelength & Frequency

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A photon's energy is determined by its frequency, expressed by the equation E = hf, where h is Planck's constant. This relationship indicates that higher frequency electromagnetic waves, such as 400 GHz, possess greater energy compared to lower frequencies like 20 Hz. The discussion also touches on the application of classical kinetic energy equations to photons, clarifying that E = 1/2mv^2 is not applicable due to photons having zero mass. Instead, the energy-momentum relationship for photons is simplified to E = pc, where p is momentum. Overall, the conversation emphasizes the unique properties of photons in relation to energy and frequency.
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How is a photon's energy determine in relation to it's wavelength and frequency?
For example, 20hz vs. 400ghz electromagnetic waves.
 
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nuby said:
How is a photon's energy determine in relation to it's wavelength and frequency?
For example, 20hz vs. 400ghz electromagnetic waves.
The energy of a photon, E (which can be considered as all kinetic energy since the proper energy = E0 = 0 and E = K + E0 = K), is related to the photon's frequency, f, by E = hf where h = Planck's constant = 6.626068 × 10-34m2kg/s.

Pete
 
can E=1/2mv^2 be applied to photons ever?
or E=mc^2
 
The proper relativistic equation is
:E^2 = p^2c^2 + m^2 c^4, which works just fine for photons when m = 0.

For ordinary particles, one can Taylor expand E = \sqrt{p^2c^2 + m^2 c^4} to get a non-relativistic equation most people use... but for photons, you can't do this, and E = pc simply.

According to de Broglie, p = h \nu, of course.
 
nuby said:
can E=1/2mv^2 be applied to photons ever?
or E=mc^2
No.

Pete
 

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