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So an electron can only orbit a nucleus where its wavelength makes a standing wave, leading to discrete energy levels in atoms. That is the Bohr model which has been superseded by quantum mechanics. In quantum mechanics the electron occupies definite energy levels which arise because of the potential well that is generated by the nucleus. As the other ...


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Well, you answered your question partly yourself. Only a finite amount of states (= energy and wavelength of the electron) are allowed, this is the basis of quantum mechanics. If an electron is in a certain state (ie. it has a certain energy and is in a certain orbital), you can then try to calculate its energy and wavefunction and from that its momentum (as ...


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Please completely abandon the idea of an electron orbiting a nucleus. That is only believed in the now long defunct Bohr model of the hydrogen atom. In modern Quantum Mechanics, in order to describe the electron in a hydrogenic (mono-electronic) atom, we need to solve the atom’s Schrödinger Equation which yields the wave functions ...


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When these terms are used, light is been pictured as an electro-magnetic wave. Thinking of the microwave example, inside what you would have is like a sea where light is equivalent to the undulations. So in this picture, the wave-length is the distance between two consecutive wave peaks. Larger wave-length implies more separation between these peaks. Also ...


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...in what instances is the kinetic energy of a free particle... Remark: The temperature dependent expression for the kinetic energy is not a property of a single particle, but an ensemble. The $\frac{3}{2}$ comes from a statistical consideration of degrees of freedom of a mechanical particle scheme. The $\pi$ comes form the wave picture. Using a ...


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here is an animated version of image for more in formation refer to http://www.physicsclassroom.com/class/waves/Lesson-3/Boundary-Behavior and a good tool for playing with https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html


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My intuition is that the frequency should stay the same because the waves in the light rope are caused by the waves in the heavy rope. The point where the ropes attach will oscillate with a common frequency. So, for $(b)$, the frequency would be the same. For $(c)$, use the equation $v= f\lambda$. You already correctly determined that the velocity ...


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Well the imaginative experiment you have put forth is a fairly satisfactory method I would say. The wavelength of a guitar string is simply the distance between two consecutive peaks or troughs of the oscillating string (for simplicity let the frequency be constant). The wavelength of the electromagnetic field, or light, is just the corresponding value for ...


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To a good approximation the radiation emitted from a hot piece of steel will be the same as emitted by a black body. The relationship between the wavelength $\lambda$ of light emitted by a black body and the temperature $T$ is given by Planck's law: $$ B = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_b T} - 1} $$ where $B$ is the spectral radiance, $h$ ...


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Typically electromagnetic radiation starts with movement of an electron, a charged particle. Either as a varying current, say in an antenna, or within an atom when an electron drops to a lower energy state and changes shell within the atom. In either case there is change in movement of the charge, which emits energy as a photon, aka a quantum of ...


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I'm going to interpret your question a bit liberally - you ask for the case where we ignore the Sun; I'm going to go a little bit further and ignore the entire galaxy (and in fact other nearby galaxies) and talk about the cosmic background radiation. The cosmic microwave background gets a lot of attention, but in fact there are cosmic backgrounds at a very ...


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EDIT: Now that the question has been reformulated, this isn't the actual answer (it gives the value of the average detected wavelength) Two assumptions: We're looking for some average values in the cosmos, not the specific value we get here on Earth (since the sun messes the whole calculation up) We're looking for the average wavelength, not the most ...


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The wavelength of sound is of the order of 1 meter. So any objet visible to the eye can deflect it. In the case of light , we need sofisticated equipment to observe the effect of diffraction. Diffraction of light an be best observed when a small slit is used. THe slit should be the order of few microns. Then the light is diffracted which is observed in the ...


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As a general guide, if we consider diffraction of a wave with wavelength $\lambda$ from an object of size $d$ then the characteristic angle of the diffraction is given by: $$ \sin\theta = O\left(\frac{\lambda}{d}\right) $$ where the $O()$ symbol means of order i.e. roughly the same as. So for example in a Young's slits experiment, where $d$ is the slit ...


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Wavelength and wavenumber are redundant terms, as it sounds like you know. Their use is a matter of convention, which in my experience changes from field to field which you won't know until you've been around. So...if you know which one people use, go with the flow. Otherwise, use which one you know, and be confident; for questions of order-of-magnitude ...


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I don't think there's much to say beyond the obvious: You should use whatever terminology is most helpful in communicating the information that you want to communicate. That has to do with the audience you're talking to. Just like how you use °F when talking to Americans and °C when talking to non-Americans ... similarly it's often wise to use cm^-1 when ...


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Traditionally wavenumber is used in molecule spectrums such as infrared spectrums in organic chemistry where it is given in the incoherent SI-unit $\textrm{cm}^{-1}$. Mostly because one obtains convenient numbers on the axis. Also in most of the wave equations it is used, because again you can make the convenient substitution $k \equiv \frac{2\pi}{\lambda} = ...



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