Density of States for Electrons in Metal: Exploring the 2

[KFC]

Consider electrons in metal as a quantum ideal gas, the quantized energy would be

$$E(n_x, n_y, n_z) = \frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)$$

Now, letting

$$\alpha = \frac{n_x}{L_x}, \qquad \beta = \frac{n_y}{L_y}, \qquad \gamma = \frac{n_z}{L_z}$$

and

$$\rho^2 = \alpha^2 + \beta^2 + \gamma^2$$

and setting up an coordinate with $$\alpha, \beta, \gamma$$. One could see that the volume b/w E-E+dE is proportional to the volume b/w $$\rho - \rho + d\rho$$

Note that the volume within $$\rho - \rho + d\rho$$ is just one-eighth of the volume of the spherical shell with thickness $$d\rho$$, namely,

$$V_\rho = \frac{4\pi \rho^2 d\rho}{8}$$

Hence, the density of states within $$\rho - \rho + d\rho$$ becomes

$$N(\rho)d\rho = \frac{V_\rho}{\textnormal{volume per state}} = \frac{4\pi\rho^2d\rho /8}{1/V} = \frac{\pi V \rho^2 d\rho}{2}$$

Well, we know that

$$E = \frac{h^2}{8m}\rho^2$$

in addition

$$N(\rho)d\rho = N(E)dE$$

which gives

$$N(E) = \frac{1}{2}\frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}$$

But the correct answer for density of state in energy should be (two times of the above reuslt), i.e.

$$N(E) = \frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}$$

Where is the '2' come from?

Someone suggests

$$N(\rho)d\rho = 2\times\frac{\pi V \rho^2 d\rho}{2}$$

This will give the correct answer, but where is "2" come form?

 

[Avodyne]

Each electron has two possible spin states.

 

[KFC]

Each electron has two possible spin states.

Oh, how can I forget that! Thanks a lot :)

 

[nsundar]

Can someone help me calculating the density of states from phonon dispersion relations.
I have computed the phonon dispersion relations from considering the vibrations of crystals. Now as a next step, i want to compute the density of states. not sure how.
many thanks in advance for your help