Formula: Free electron gas in 1d (density of states) Density of states Energy
$$D(W) ~=~ \frac{L}{\pi} \, \left(\frac{2\class{brown}{m}}{\hbar^2}\right)^{1/2} \, \frac{1}{\sqrt{W}}$$
$$D(W) ~=~ \frac{L}{\pi} \, \left(\frac{2\class{brown}{m}}{\hbar^2}\right)^{1/2} \, \frac{1}{\sqrt{W}}$$
Density of states
$$ D(W) $$
Density of state of an electron gas - with free electrons which do not interact with each other and which are partially confined in a potential. In this case, the electron gas is confined by a potential, so it can only be in a one-dimensional line. This can be a nanowire, for example.
The density of states represents the states per energy interval, in this case for both spin directions of the electron. To get the density of states for only one spin direction, you have to multiply by \(\frac{1}{2}\). And, to get the density of states \(g(W)\) per energy interval AND per length, multiply by \(\frac{1}{L}\): \[ g(W) ~=~ \frac{1}{\pi} \, \left(\frac{2m}{\hbar^2}\right)^{1/2} \, \frac{1}{\sqrt{W}} \]
Length
$$ L $$ Unit $$ \mathrm{m} $$
Length of a one-dimensional solid (for example, a nanowire) in which the electron gas is confined.
Mass
$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$
Mass of a Fermi particle which is in the Fermi gas. In the case of an electron gas, it is the (effective) mass of the electron.
Energy
$$ W $$ Unit $$ \mathrm{J} = \mathrm{Nm} = \frac{ \mathrm{kg} \, \mathrm{m^2} }{ \mathrm{s}^2 } $$
Energy of a state which the electron can occupy. In the one-dimensional electron gas, there are fewer states at higher energies because of \( D \sim \frac{1}{\sqrt{W}}\).
Reduced Planck's constant
$$ \hbar $$ Unit $$ \mathrm{Js} = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{s} } $$
Reduced Planck's constant is a physical constant and has the value:
$$ \hbar ~=~ \frac{h}{2\pi} ~=~ 1.054 \, 571 \, 817 \cdot 10^{-34} \, \mathrm{Js} $$