Density of States
It is defined as total number of available electronic states per unit volume per unit energy. It is denoted as z(E).
To calculate the number of energy states with all possible energies, with n as radius (where n = n12 + n22 + n32) a sphere is constructed in three dimensional space shown in Fig.1. Since n1, n2 and n3 can have only +ve integer values, we have to consider only one octant of the sphere. The number of energy states within a sphere of radius n,
\[ \frac{4}{3}\pi n^3 \tag 1 \]The available energy states in one octant of the sphere,
\[ \frac{1}{8}\times \frac{4}{3}\pi n^3 \tag 2 \]In order to calculate the number of states within a small energy interval E and E+dE, we have to construct two spheres with radii n and n+dn, calculate the space occupied within these two spheres.
Figure 1. a) Energy states within the n and n+dn radius of the sphere. b) 1/8th of the sphere for positive integer
Thus the number of energy states having energy values E and E+dE.
\[ N(E) dE = \frac{1}{8} \frac{4\pi}{3} (n+dn)^3 - \frac{1}{8}\frac{4}{3}\pi n^3 \tag 3 \] \[ N(E) dE = \frac{1}{8} \frac{4\pi}{3} \left[ n^3+dn^3+3n^2dn+3d^2n-n^3 \right] \tag 4 \]Neglecting high power of dn as it is very small,
\[ N(E) dE = \frac{1}{8} \frac{4\pi}{3} \left[ 3n^2 dn \right] \tag 5 \] \[ N(E) dE = \frac{\pi}{2} \left[ n^2 dn \right] \tag 6 \]The expression for the energy of electron is given by
\[ E = \frac{n^2 h^2}{8mL^2} \tag 6 \] \[ n^2 = \frac{8mE L^2}{h^2} \tag 7 \] \[ n = \left[ \frac{8mE L^2}{h^2} \right]^{1/2} \tag 8 \]Differentiating Eqn 7 on both sides,
\[ 2n \, dn = \frac{8mL^2}{h^2} dE \] \[ dn = \frac{1}{2n} \frac{8mL^2}{h^2} dE \tag 9 \]Substituting the value of \( n \) from Eq. (8) into the above equation,
\[ dn = \frac{8mL^2}{h^2} \cdot \frac{1}{2} \left[ \frac{h^2}{8mL^2} \right]^{1/2} \frac{dE}{E^{1/2}} \] \[ dn = \frac{1}{2} \left[ \frac{8mL^2}{h^2} \right]^{1/2} \frac{dE}{E^{1/2}} \tag {10} \]Substituting values of \( n^2 \) and \( dn \) from Eq. (5),
\[ N(E) dE = \frac{\pi}{2} \left[ \frac{8mL^2 E}{h^2} \right] \cdot \frac{1}{2} \left[ \frac{8mL^2}{h^2} \right]^{1/2} \frac{dE}{E^{1/2}} \] \[ N(E) dE = \frac{\pi}{4} \left[ \frac{8mL^2}{h^2} \right]^{3/2} E^{1/2} dE \tag {11} \]According to Pauli’s exclusion principle, two electrons of opposite spin can occupy each state and hence the number of energy states available for electron occupancy is given by
\[ \begin{align} N(E) dE &= 2 \times \frac{\pi}{4} \left[ \frac{8mL^2}{h^2} \right]^{3/2} E^{1/2} dE \\ N(E) dE &= \frac{\pi}{2} \left[ \frac{8mL^2}{h^2} \right]^{3/2} E^{1/2} dE \\ N(E) dE &= \frac{4\pi}{h^3} (2m)^{3/2} E^{1/2} dE \times L^3 \tag{12} \end{align} \]Density of states is given by number of energy states per unit volume,
\[ Z(E) dE = \frac{N(E) dE}{V} = \frac{4\pi}{h^3} (2m)^{3/2} E^{1/2} dE \tag{13} \] Where \( \text{V}=L^3 \)