Consider electrons in metal as a quantum ideal gas, the quantized energy would be(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

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)

[/tex]

Now, letting

[tex]

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

[/tex]

and

[tex]

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

[/tex]

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

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

[tex]V_\rho = \frac{4\pi \rho^2 d\rho}{8}[/tex]

Hence, the density of states within [tex]\rho - \rho + d\rho[/tex] becomes

[tex]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} [/tex]

Well, we know that

[tex]E = \frac{h^2}{8m}\rho^2[/tex]

in addition

[tex]

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

[/tex]

which gives

[tex]

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

[/tex]

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

[tex]

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

[/tex]

Where is the '2' come from?

Someone suggests

[tex]N(\rho)d\rho = 2\times\frac{\pi V \rho^2 d\rho}{2} [/tex]

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

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# Density of state

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