Hi Tian, unfortunately the number of characters is quite limited so it's easy to blend things that do belong to different contexts but are represented by the same character. That's here the case with "E".
E in E=hv represents energy, and the equation yields the energy of one quantum for given v (= "ny", the frequency). If you had monochromatic radiation with known power and known frequency, you could divide the power by the quantum energy and would get the number of photons per second in this radiation.
E in electromagnetic radiation denotes the strength of the electric component of the radiation. The power flow density is given by the Poynting vector S = E x H. In order to calculate the energy of a pulse you must not only know the values of E and t but the area of the cross-section of the pulse. Your first example: In vacuum, wave impedance Z0 = 377 Ohm; Ex = 10 V/m, Hy = Ex/Z0, direction of propagation toward positive z-values, area A = 1 m^2, t = 1 s =>
power flow density S = E^2 / Z0 = 100 / 377 W/m = 0.265 W/m
power P = S * A = 0.265 W/m * 1 m = 0.265 W
energy W = P * t = 0.265 W * 1 s = 0.265 Ws
Your second example: as the first, but Ex = 5 V/m gives only a quarter of S in the first example, and thus only W = 0.0663 Ws.
So, the answer to your main question is: Not at all, because E=hv is not restricted to electromagnetic phenomena. I hope that will help.
Hi Tian,
the expression:
E= hv,
gives you the energy of a single photon. Try to understand your question from the quantum nature, where higher the energy, more the number of packets. Hope this helps.
Dear Tian Seng, since the energy of a single photon is very small, the fields of your cases are composed of billions of photons per m^2 per s. Yes, the higher the photon count the brighter the light while the energy per photon depends on the color of the light.
You might ask "How strong is the E field of a single photon of energy E?" but the answer would at best be the result of an approximation because one cannot measure the field components of a single photon, and if someone claims to know exactly the diameter and the length of a photon (which would be needed to arrive at an exact result) then he is probably mistaken.
I guess if you google "photo-electric effect Einstein 1905" you might find a description of the basic problems. Wishing you success
Dear Tian Seng Chan, please follow the link:
https://en.wikipedia.org/wiki/Photoelectric_effect
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