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7

All photons are excitations of the photon quantum field. The photon is actually not a great description of propagating light - it tends to be far more useful when you're considering light exchanging energy with something else. If you insist on a photon description then you'd have to treat the radiation as a superposition of photons. A beam of light would be ...


7

Is it possible the: is ke meaning kinetic energy? I ask because the equation for the de Broglie wavelength of a non-relativistic particle is: $$ \lambda = \frac{h}{\sqrt{2m_0 KE}} $$ where $KE$ is the particle's kinetic energy.


3

Avoiding mathematical formulae to the maximum, and warning for furious hand-waving ahead, I would state it like this: In the classical picture, there is no quanta concept, so you could have just a little bit of radiation energy at any frequency. However, since quantization appeared, the minimum amount of radiation energy that you could possibly have at a ...


2

You'll need two things, the solar zenith angle $\theta$ and the air mass $AM$. The zenith angle is given by $$\cos\theta = \cos\phi \cos \delta + \sin\phi \sin\delta \sin h$$ where $\phi$ is your latitude, $\delta$ is the declination of the Sun, and $h$ is the hour angle. You should know your own latitude. The declination of the Sun can be found on a ...


1

"Bright light can never hurt your eyes" seems false to me… enough energy focused on the retina will cause damage, regardless of the wavelength. Otherwise you would not need to wear laser goggles… That aside, materials typically have certain ranges where they absorb light more strongly than others. There is no hard and fast rule for this, but if you google ...


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Your curve has two components in it: Actual solar intensity per unit area (affected by atmospheric conditions, that is the probability that the sunlight is not scattered or absorbed on its path through the atmosphere) The projection of the sunlight onto your collector: if your collector is facing in a particular direction, the angle of the normal of the ...


1

If the light passes first through medium $\tau_1$, then through medium $\tau_2$, the optical depths add up to $\tau = \tau_1 + \tau_2$ , as you can see from applying the Beer-Lambert-Formula twice. Does it matter whether the media overlap (if the media don't interact and light propagation is linear)? No, it doesn't. The optical depth comprises the ...


1

I think you misunderstand the ultraviolet catastrophe. It does not mean that the energy radiated reaches zero at any finite frequency, just that the power tends to zero as the frequency tends to infinity. Pre-quantum physics thought that blackbody radiation was ruled by the Rayleigh-Jeans law $$ B_\nu(T) = \frac{2\nu^2 k T}{c^2} $$ This does obviously not ...


1

Could anyone explain how this process complies with the second law of thermodynamics? The Stefan-Boltzmann law. I'll start with an ideal black body. Black bodies absorb all incoming radiation. They also emit radiation as a function of temperature. The peak frequency and the intensity increase as temperature increases. The emitted power is given by the ...


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Let's suppose the broadening mechanism is van der Waals or Stark broadening - something where the energy levels of individual atoms are perturbed. In this case you could use the following argument. Divide the line profile up into groups of atoms which share the same perturbation and treat each of these as a subpopulation with a different energy gap and ...



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