qno6 calculate the density of state function for a two dimensional gas

Question

Q.NO.6 Calculate the density of state function for a two-dimensional gas of independence particles, using the methods developed for three - dimensional case. assume that gas is confined by a rectangular container of area A whose sides are Lx and Ly . The density state function is a reflection of how energy levels are spaced and how many quantum states are allotted to them. These quantities are determined ultimately from Schrodinger equation and the boundary conditions , and thus depend only on the dynamic properties of the system. The density of states is therefore a dynamical quantity and is independent of any probabilities phenomena consider nth energy level of energy En to which there belong qn distinct quantum states, the number qn is referred to as the degency factor of the level. The average number of particle Nn occupying this level can be written Nn = qn F(En) ---(1) It is clear from two above expressions that the degency factor and the density of states functions are essentially similar quantities . The distribution function and the density of states or the degeneracy , we can calculate the average of any energy- dependent parameter E <\xi>=\frac{\int \xi\mleft(\varepsilon\mright)N\mleft(\varepsilon\mright)d\varepsilon}{\int N\mleft(\varepsilon\mright?d\varepsilon}\mleft.=\frac{1}{N}\xi(\varepsilon)N\mleft(\varepsilon\mright?\mright?d\varepsilon In three - dimensional , phase space has six coordinates x,y,z,px,py,pz . Since the above equations are independent of the coordinates x,y and z we can plot our result in a three-dimensional momentum space (px,py,pz) without loss of generality . In space, the above equation defines a spherical surface of radius p or "(2mE)^1/2" fro a free particle of constant energy "E". For a slightly greater energy "E=dE" corresponding to a large momentum p+dp to a large momentum p+dp the constant - energy sphere will increase in radius by an amount dp. It is clear that relationship between dp and dE is pdp = mdE ----(2) . The allowed momentum eigenvalues given by eq(2) can be plotted in this diagram as a rectangular orthogonal lattle of points correspond to all possible integral values of quantum number nx,ny and nz. In such a lattic , the "unit cell" is a rectangular volume as illustrated in figure . For this it is possible to determine how many quantum states there are in any energy range dE- which is by definition the density of states-simply by calculating the phase space volume between two spherical shells shown in diagram
The volume of momentum Space per quantum state \left.D P_{x} \Delta P y D P z=\frac{h^{3}}{2 x \operatorname{ly} L_{2}}\right] \rightarrow 63 If the particles have spin =1/2 the quantum states will have spin degency in which two states of opposite spin ms=+-1/2------(8) The density of states is now multiplied by a factor of two the result being q(\varepsilon)d\varepsilon=\frac{8\sqrt{2}\pi v}{h^3}m^{3/2}\sqrt{\varepsilon}d\varepsilon This can be recognized as a relationship stemming from Heisenberg uncertainty principle