I know that the energy of one photon equals $E=hf$ and that concept of photons came from the photoelectric effect. My question here can we assume that a photon equals one wavelength for example if we have a light with wavelength equal to $700\,\mathrm{nm}$, one photon of that light will have energy $=2.84 \times 10^{-19} \mathrm{J}$. Does that mean if we sent the same kind of light that has two wavelengths of $700\,\mathrm{nm}$, it will have the energy of two photons?
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1$\begingroup$ Can you explain what you mean by "if we sent the same kind of light that has two wavelengths of $700\,\mathrm{nm}$, it will have the energy of two photons"? $\endgroup$– SchrodingersCatDec 19, 2015 at 17:11
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$\begingroup$ i mean we used the same light that has wavelength 700nm but we only sent 2 waves of it. sorry for the bad english $\endgroup$– Omar AliDec 19, 2015 at 17:12
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$\begingroup$ A "wavelength" is not a thing. It is a property of light. $\endgroup$– HDE 226868Dec 19, 2015 at 17:13
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$\begingroup$ Do you mean you want to send light wave that has completed exactly 2 time periods? $\endgroup$– ShankRamDec 19, 2015 at 17:17
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$\begingroup$ ya, that has two troughs and two crests $\endgroup$– Omar AliDec 19, 2015 at 17:51
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3 Answers
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can we assume that a photon equals one wavelength for example if we have a light with wavelength equal to 700nm, one photon of that light will have energy =2.84×10−19J.
No. A photon is a quantum of energy and the no. of wave periods you have in a pulse is not related to the no. of photons present in that wave pulse. It depends on intensity. The more intense the light pulse is the more no. of photons it will have and the energy carried by one quanta of that light pulse will be $hf$. Photons are point like particles and it's not helpful to think of them being a one complete wave period.
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As others stated, the photon is an elementary particle and has no extent in space, so the wavelength characterizing it is connected with the wave train it is a member of.
A classical wave has a wavelength that just labels the photon and allows for the $E=h\nu$ observation of its energy. The classical wave is a continuous train of lambda wavelengths built up of zillions of photons.
A huge number of photons traveling with velocity $c$ of energy $h\times f$ build up each individual lambda in space . If you know the amplitude of the wave, E for example, which gives the Poynting vector, you can then know the energy within a lambda length and can calculate how many photons build it up. This link shows how mathematically this coherent build up can happen .When you learn the rudiments of Quantum Electrodynamics you will be able to understand it. It shows how photons build up the classical electromagnetic wave.
This continuity between quantum and classical happens because the quantum mechanical wave function of the photon is a solution of a quantized form of Maxwell's equations.
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You are trying to reconcile the quantum mechanical and electromagnetic definitions or properties of light, which is quite unintuitive and difficult at best.
Let me help you understand some things.
For a particular observation of a phenomenon related to light, you either treat it as a wave or as photons. Never both together. You can relate the intensity of a wave to the number flux of photons per unit area, but for some calculations, combining the 2 models results in impossible methods for, out indeterminate solutions.
Coming back to your question, a particular photon can be treated not as a wave period, but as a "wave packet", a superimposition of a number of waves centered at a particular wavelength (dependent on the medium of propagation) and frequency(determined by the photon source and independent of the medium of propagation). The photon's kinetic energy can be said to be determined by this centering frequency, as $E=hf$.
If you say 2 waves, then the observed energy will depend on your position, the orientation of the waves, and half a dozen other things. I refer here to interference, diffraction and superposition of waves, which lead to energy redistribution between different observer positions.
If you say 2 photons, and take them to be incident at the detector either simultaneously or serially, the net energy recorded will indeed be the sun of the energies of the 2 individual photos incident on the detector.
answered Dec 21, 2015 at 4:07