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1

About the limit: $\frac{\sin[\frac{\pi}{d}(1/N+1)x]}{\sin[\frac{\pi x}{dN}]}= \sin[\frac{\pi}{d}(1/N+1)x]\times \frac{\frac{\pi x}{dN}}{\sin[\frac{\pi x}{dN}]}\times \frac{dN}{\pi x}\approx \sin[\frac{\pi}{d}x] \times 1 \times \frac{Nd}{\pi x}$ in the last step I used $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $1/N+1\approx 1$. after rearranging the terms it ...

0

The expected effects at elementary level would be that the positive gate potential would attract the electrons into the well, and also that it would counteract the built -in field orthogonal to the surface. Orthogonal field reduces mobility, so reducing it increases mobility. This elementary view may not, of course, be correct.

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When the atoms get close to each other, the electrons of the first begin to feel the nuclei of the second, and vice versa. So, for example, when two hydrogen atoms approach each other, we have to consider the the potential due to all four particles (and the Pauli principle, but that issue doesn't have a bearing on the kernel of your question). The ...

0

No the energy band would not reverse. As we increase the applied volatge the drop across the nuentral region increases such that the n side depletion region is always higher in potential to the p side depletion region. Actually : Depletion potential = (built in potential) -{(applied potential)-(potential drop across neutral region) } But at lower applied ...

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If you just take the empty bandstructure, you will see that any periodic arrangement of atoms (conductors, semiconductors, insulators) features a set of allowed bands and forbidden regions, so called bandgaps. Fully occupied bands can not contribute to electrical current. There are no free places, where carriers could move. Only partially occupied levels ...

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An example of a doped semiconductor might give an intuitive picture of some aspects of this topic: Consider a material like Germanium. Atoms are structured in a lattice. All valance electrons are "used" in the crystal structure to form bonds to neighbours; none are more "free" than others. source Now dope it with another atom of one higher electron ...

12

No band is special. A partially full valence band does conduct, just like a partially full conduction band. On the other hand, a perfectly full band conducts just as well as a perfectly empty band: No conduction at all. Now, nobody is surprised when you say an empty band can't conduct, but at first it seems surprising that a full band is the same way. ...

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A band is essentially a (near) continuous collection of momentum eigenstates. Within the band the electrons can be treated as free to a reasonable approximation, so their eigenstates are just plane waves. The symmetry means that for every eigenstate there is another with an equal and opposite momentum. So if we populate every momentum eigenstate the net ...

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$$g(E)=\text{number of states at energy E available to be occupied}$$ $$f(E)=\text{probability that a state with energy E is occupied}$$ so that $$g(E) \ f(E) = \text{average number of occupied states with energy E} \\ =\text{average number of particles with energy E} = N(E)$$ So that the total number of particles will be given by $$N=\int N(E) \ d E$$ ...

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