KFC
Dec18-08, 03:40 PM
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?
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?