# Tag Info

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I think they are probably trying to characterize small mechanical devices. One part of this analysis would be analyzing their motion. One motion it can do is to vibrate. If the wavelength of the vibration is much smaller than the size of the object, then the vibration can't tell how small the object is, and the vibration will behave the same way it would ...

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Polymers come in many different shapes and forms in terms being crystalline or amorphous. Broadly speaking we can categorise them as follows. 1. Semi-crystalline thermoplastic polymers: like polyethylene (PE). Extremely high Molecular Mass (100,000 and higher) make these very long strands difficult to align into crystallites. Relatively small areas in the ...

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A general highly simplified overview: Generally macromolecules/polymers exhibit very complex structures. The completely disordered/amorphous state is on the one hand to be expected from purely configurational entropy considerations, on the other hand there are cases where the chains or side-chains would energetically favor to be aligned, but this can almost ...

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The Fermi energy is the energy of the highest occupied state at absolute zero of a system of non-interacting (or mean-field interacting) fermions. So for a band structure, the Fermi energy is the energy of the highest-occupied level after you have filled al your bands with electrons. Note that each (spin) band can hold $N_{\textrm{cells}}$ electrons, where ...

0

In STM, you tunnel from states of your tip into states of the sample. Electrons can tunnel into states within the whole bulk and the matter beneath. However, their spectral weight decays rather quickly. As a rule of thumb in STM, your current increases by one order of magnitude per Angstroem that you reduced the tip-sample distance. So, the DOS of matter 1 ...

5

Metals are good conductors of electricity because the outer (valence) electrons of the metal atoms are only loosely bound to the nucleus and form molecular orbitals known as the conduction band. Electrons can move more or less freely through the conduction band and so metals conduct electricity generally well. When a metal is chemically oxidised its outer ...

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It's because valence electrons are bounded. For example, consider Si and SiO2. While Si is semiconductor, SiO2 is insulator because it has no free valence electrons. BTW, many metal oxides ARE NOT in fact insulators - for example ZnO, Fe3O4 are all conductors. But it's true that oxides of metals have lower conductivity than pure metals.

1

I would say the maths and equations are pretty much identical except in H NMR you would use the gyromagnetic ratio for a proton, while in EPR you use the data for an electron. Both are spin 1/2 systems. In terms of medical imaging it is easier to pick H2O via pulse NMR (rather than continuous field i.e what chemists do for molecules etc) than observe free ...

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After x-rays hit a substance they will be scattered in all directions; if the material is a crystal then you will obtain a diffraction pattern where each point is created by the constructive interference of the scattered rays. The connection between the diffraction pattern and the reciprocal space is readily found: take a crystal and consider an atom ...

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I don't believe that the thermal conductivity of most metals is very sensitive to magnetic fields. Yes, there will be some field-induced band shifting in the case of an itinerant ferromagnet which, in principle, leads to a change in the density of states at the Fermi level, but that will typically be a very small effect. If the magnetic field induced ...

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Magnetic fields certainly can influence thermal conductivity. This shows up, not surprisingly, when there is a strong influence of the magnetic field on other properties, particularly electronic ones. One (non-metal) example is 'Thermal conductivity tensor in YBa$_{2}$Cu$_{3}$O$_{7-x}$: Effects of a planar magnetic field' by R. Ocana and P. Esquinazi, Phys ...

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Since the question is rather vague, I will just give you some key points: Debye's model treats oscillation modes of a solid as sound waves (phonons) with frequency $\omega(\mathbf{k})=v|\mathbf{k}|$ ($v$ the sound velocity). As a result, with this model, Debye shows how the heat capacity is directly related to the rate of change of the energy expectation ...

1

When a liquid or solid evaporates, it turns into a gas. In a closed container, pressure builds as gas accumulates. There are two competing processes. In the solid or liquid, the higher energy atoms at the surface fly off. In the gas, the slower atoms stick to the surface and condense. The number of atoms available to condense is proportional to the gas ...

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Your answer is correct! The answer given in Wikipedia is also correct! The discrepancy in the requirements of (h,k,l) comes in your choice of the reciprocal lattice(RL) basis vectors. In Wikipedia, the RL basis vectors are derived from the non-primitive direct lattice basis vectors a1=ax, a2=ay,a3=az. Thus the (h,k,l) are the coefficients of the ...

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Density Functional Theory (DFT) is used to calculate the electronic structure and properties of metals as much and "successfully" as it is used for molecules, clusters, alloys, insulators and semiconductors. Of course there are certain things that DFT is good at and can and cannot do. However saying that "DFT calculations are not accurate for metallic ...

-1

DFT can theoretically depict the ground state of any material, however, we can only deal with the properties of material by using some approximation. the quality of your approximation determines the degree of accuracy.

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The dopant level is bound to the atom, and we can think of two different "ionization energies" that exist: one would be to freedom in the semiconductor's conduction band (call this $E_{d}$ - what we are interested in), and a larger one is the energy to complete freedom in the air (call this $E_{free}$). Electrons in the semiconductor can also be excited to ...

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If we de ionise sun it would re ionise itself into older star due to huge gravity due to mass. we can only stop its fusion in its some parts and remove them which would give us gaseous hydrogen which we have to liquefy in low pressure and temperature

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Caesium From: "Physics 1942 – 1962: Including Presentation Speeches and Laureates' Biographies" by Yong Zhou

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These lectures on the QHE and FQHE given at the Les Houches summer school are a great start imho.

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Fermi surface is not necessarily a sphere. It can have an arbitrary shape in the reciprocal space (momentum space) that respects the symmetry of the crystal. Because crystals are periodic arrangements of atoms in direct space, in the reciprocal space it implies that the Fermi surface must repeat itself periodically in every direction. Every such "period" is ...

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I think the problem is not in the calculational steps above, but in the very first expression, the ‘two-electron integral’, which seemingly aims to calculate the expectation value of the Coulomb interaction in terms of some (non-interacting) Bloch wave-functions, $\phi_{n{\bf k}} ({\bf r})$. However, the notation seems to be incorrect, since the ...

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In general solids can store more energy in their electric fields due to their polarization mechanisms. "Permittivity (denoted by ε; measured in farads/meter, F/m) is the property that permits a substance to store energy in, and release energy from, an electric field. This property allows a substance to buffer any change in the applied electric field. The ...

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The quantity $\langle c_{i\sigma}^{\dagger}c_{j\sigma}\rangle$ is in general not real and positive. Hence the contribution of the hopping term to the total energy doesn't necessarily become more negative as more particles are added. Note also that to be precise, the left-hand side of Eq. 1 should be $H-\mu N$, where $N$ is the number operator. Many people ...

2

For the question of why a deformation of a bouncing ball would convert some of the energy of the system into heat energy, you need to think about what heat actually is in the ball. Heat will be transferred to the ball in the form of vibrations of the atoms that make up ball. When a collision occurs, the ball compresses, and upon restoration, there will be ...

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Provided that you have been provided or can calculate the coefficient of restitution $e$ for the collision, you could find $K_i$ the kinetic energy before the collision and $K_f$ the kinetic energy after the collision. Then the part which was transferred into heat is: $$H=\Delta K=K_f-K_i$$ And as a percentage, $$\% H=\frac {\Delta K}{K_i} * 100 \%$$

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