Question

Obtain an expression for energy levels and density of states in one dimension.​

Answer 1

Answer:

The density of states in a semiconductor equals the density per unit volume and energy of the number of solutions to Schrödinger's equation. We will assume that the semiconductor can be modeled as an infinite quantum well in which electrons with effective mass, m*, are free to move.

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Answer 2

Answer:

When a large number of neighboring orbitals overlap, bands are formed. However, the natures of these bands, their energy patterns, and their densities of states are very different in different dimensions.

Before leaving our discussion of bands of orbitals and orbital energies in solids, I want to address a bit more the issue of the density of electronic states and what determines the energy range into which orbitals of a given band will split. First, let’s recall the energy expression for the 1 and 2- dimensional electron in a box case, and let’s generalize it to three dimensions. The general result is

E=∑jn2jπ2ℏ22mL2j(2.3.1)

where the sum over j runs over the number of dimensions (1, 2, or 3), and Lj is the length of the box along the jth direction. For one dimension, one observes a pattern of energy levels that grows with increasing n , and whose spacing between neighboring energy levels also grows as a result of which the state density decreases with increasing n . However, in 2 and 3 dimensions, the pattern of energy level spacing displays a qualitatively different character, especially at high quantum number.

Consider first the 3-dimensional case and, for simplicity, let’s use a box that has equal length sides L . In this case, the total energy E is ℏ2π22mL2 times (n2x+n2y+n2z) . The latter quantity can be thought of as the square of the length of a vector R having three components nx , ny , nz . Now think of three Cartesian axes labeled nx , ny , and nz and view a sphere of radius R in this space. The volume of the 1/8 th sphere having positive values of nx , ny , and nz and having radius R is 18(43πR3) . Because each cube having unit length along the nx , ny , and nz axes corresponds to a single quantum wave function and its energy, the total number Ntot(E) of quantum states with positive nx , ny , and nz and with energy between zero and E=(ℏ2π22mL2)R2 is

Ntot=18(43πR3)=18(43π[2mEL2ℏ2π2]3/2)(2.3.2)

The number of quantum states with energies between E and E+dE is dNtotdEdE , which gives the density Ω(E) of states near energy E :

Ω(E)=dNtotdE=18(43π[2mEL2ℏ2π2]3/232E−−√).(2.3.3)

Notice that this state density increases as E increases. This means that, in the 3-

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