10

If we speak of electrons, then it is appropriate to speak of "wind* as a flow of atoms. Electrons are usually confined to atoms or materials, inside of which the wind atoms do not penetrate, so the collision of electrons with wind are very unlikely. One place where one does encounter a stream of unbound electrons is electron microscope. In this case one ...


6

If one imagines a macroscopic object moving through air, one can very well approximate air resistance as the action of a continuous fluid. At low speeds, this situation can be described by viscous friction, which is a force proportional to the speed of the object, and with direction opposite to its direction of motion (I neglect here the effect of turbulence)...


3

These are two different regulators. The first is designed to act on coordinate space N-body wave functions, and the second one can be used in N-body perturbation theory (or on the lattice). However, they are constructed to do the same thing, to express the coefficient of a delta function potential in terms of the physical scattering length. We are ...


3

Both methods are closely related to each other, in the sense that they describe the same low-energy (long-distance) physics. (As long as the s-wave scattering length $a$ is the same, of course.) For dilute gases at low enough temperature, the physics is universal, in the sense that it is independent of the microscopic details of the potential: it only ...


2

First, while your textbook may assume that the target atoms remained at rest (it makes the math easier), Rutherford, Geiger, and Marsden knew how to do the full kinematics problem and did not assume that. For gold and platinum it makes almost no difference in the analysis since those elements are ~50x more massive than an $\alpha$ particle (He ion). Second, ...


2

In elastic scattering one photon remains one photon. If you have an incident stream of photons (as is the case in most such experiments), then you will get scattered photons in all the directions. However, thinking of photons as classical particles is rather misleading - mathematically they arise as excitation states when quantizing electromagnetic field, ...


2

Congrats! (joke) You have discovered what's known in Astronomy & adaptive optics circles as "The shower-curtain effect." What you're observing is that the paper is a diffuse transmitter, meaning the light from the object beneath (your textbook) is transmitted but with some angle-scattering. Now, consider this as part of an optical system. ...


2

Electrons do have a mass ($9.10938356 \times 10^{-31}\,\mathrm{kg}$). The effects of this mass in your particular situation are likely immeasurable though. I partially agree to Vadims answer, collision of electrons with wind are very unlikely indeed. There is another kind of resistance electrons would experience in this case. The air that typically makes up ...


1

While proving the inequality, we can ignore the $x$ depedence of the variables, since the inequality must be satisfied for any $x$. We want to show that: $$\frac{1}{4}\leq\frac{F_2^{\rm en}}{F_2^{\rm ep}}\leq 4$$ This is equivalent to showing that $f(s,u,d)=F_2^{\rm en} - 4 F_2^{\rm ep} \leq 0$ $g(s,u,d)=F_2^{\rm en} - \frac14 F_2^{\rm ep} \geq 0$ where $s,...


1

It depends on what the wave is scattered from. The simplest case to study (as an exercise to get intuition) is scattering of a monochromatic EM wave in vacuum from an infinite metal plane: in this case the boundary condition is that the component of the electric field along the surface should be zero. This may add a phase of $\pi$ to one or both components ...


1

I couldn't understand your question properly but I think may be you're asking in one dimensional potential well problem why we don't consider the case where x tends to infinity. If that's the question then , you need to see the fact that out side the well boundary, i.e., x>l the potential is infinite and thus the particle inside the box isn't capable of ...


1

I agree with the comments to the question that using one atom layer as wings might be problematic. However, I still want to make a few points in a direct answer to the question. Carbon can adopt multiple crystalline forms, of which the most well-known as diamond, graphite/graphene, fullerenes and carbon nanotubes. Since the question is about a "sheet ...


1

This has to do with the definition of the transmission coefficient $T$ in quantum scattering experiment. What we want to know is how a particle will behave when it's "shot" at the barrier. In a general situation, part of the wave function probability will reflect off the barrier, and part will be transmitted. However, for your potential, as $t\...


1

Filling in the $V$, as given in the problem (the "one step up $V$"), in the Schrödinger equation, you'll get the wavefunction $ψ_2(x)=e^{αx}$ for the positive values of $x$. This means there is a quantum mechanical probability to find a particle with less kinetic energy than the height of the potential ($E\lt{V}$). The classical transmission ...


1

The electron mean free path $\lambda_e$ is defined in terms of the total electron collision frequency $\nu_e$, which is not a measure of the average time between collisions. Instead, it is the average time required for an electron to scatter through an angle of $90^\circ$. That is, because Coulomb collisions are long-range, the vast majority of collisions ...


1

So, why does an electron need to travel at least a distance corresponding to the mean interatomic distance (or equivalently, the ion sphere radius), before it can be scattered? So in response, off of what would an electron scatter between atoms? I think the idea is that the amplitude of phonons cannot be larger than the interatomic separation for the ...


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