# What is the dimension/unit of the density of state in this case?

My 10th edition of "Fundamentals of physics" says that by multiplying the right side of the following equation with $e^{3/2}$, we get the quantity in the unit $\mathrm{m^{-3}\cdot eV^{-1}}$: $$\rho(E) = \frac{4\pi(2m)^{3/2}}{h^3}\sqrt{E}$$

Well, I don't see how the dimensions match the unit after this multiplication. Is this even correct?

• What equation are you talking about ? Your link don't talks about units... – Spirine Oct 24 '16 at 10:13
• Sorry for not using mathjax.I have edited the question. – Mockingbird Oct 24 '16 at 10:23

$1\,eV$ is, by definition, the kinetic energy gained by an electron initially at rest, when accelerated by a $1\,V$ potential difference, which leads to $1\, eV = (1\,e)\cdot(1\,V)$.

We find experimentally that $1\,eV \approx 1.602,176,565\times 10^{-19}\,J$, and $e \approx 1.602,176,620\times 10^{-19}\,C$, so we usualy say that $1\,eV = (1e)\,J$, which is concede is, considering the dimensions, totally false. However, here we don't use $e$ as though it was a charge, but only as a dimensionless quantity, to convert energy from Joule to electron-volt.

You want to express $\rho = \frac{4\pi (2m)^{3/2}}{h^3}\sqrt{E}$ in $m^{-3}\cdot eV^{-1}$, so you must convert $E$ in $eV$ and $h$ in $eV\cdot s$. Let $E'$ and $h'$ denote the value of the corresponding quantities. We have $h' = e h$ and $E'=eE$, so

\begin{align} \rho' &= \frac{4\pi (2m)^{3/2}}{h'^3}\sqrt{E'} = \frac{4\pi (2m)^{3/2}}{(eh)^3}\sqrt{eE}\\ &= \rho\cdot e^{-3/2} \end{align}

Finally, $$\rho = \frac{4\pi(2m)^{3/2}}{h^3}\sqrt{E}e^{3/2}$$

with $m$ in $kg$, $h$ in $eV\cdot s$, $E$ in $eV$ and $e\approx 1.602$.

• Shouldn't we multiply it by only 1.6^10-19 to convert it m^-3eV-1? – Mockingbird Oct 24 '16 at 10:38
• Isn't h'=h/e? I mean 1J=1eV/e then it should be h'=h/e and E=E/e and I don't think it would be e^-3/2, but e^-5/2. – Mockingbird Oct 25 '16 at 10:56