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How do I find the dimensions of this quantity (in $cgs$)... $$\frac{4\pi me^2}{h^2n_o^{1/3}}$$

where $m$ is the mass of electron

$e$ is the magnitude of electronic charge

$h$ is the Planck's constant &

$n_o$ is equilibrium number density of electrons

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Hello zara and Welcome to Physics.SE. Generally, we guys use TeX commands for writing formulas which is one of the useful resources of SE. So from now on, Please use TeX while writing formulas for better understanding... Special courses are not required..! :) –  Waffle's Crazy Peanut Sep 27 '12 at 14:08
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1 Answer 1

There are various extensions of CGS to electromagnetism. Because $e^2$ appears in your expression without any speed of light etc., most likely, the electrostatic units – the most widespread ones – were used. In those units, the constant in the Coulomb's law known as $1/4\pi\epsilon$ in the SI units is set to one.

Equivalently, $e^2$ in our expression would get translated to $e^2/4\pi\epsilon$ in the SI units which is "electrostatic energy" except that $1/r$ is missing, so the units of $e^2$ are those of "energy times distance". However, there is one more factor of distance, $1/n_0^{1/3}$ (its inverted inverse volume, i.e. volume, to the power of 1/3, i.e. distance), so we have "energy times distance squared".

There is an additional factor of $m/\hbar^2$. Now, let's calculate the powers of centimeters, grams, and seconds. Grams cancel: one energy, one mass in the numerator; two Planck's constants in the denominator. The rest is "centimeter squared over (centimeter squared over second) squared" = "squared second over squared centimeter".

So the expression has the same units as $1/v^2$ where $v$ is a velocity. If some other CGS extension was used, it's plausible that the right units are another power of the velocity and the expression may even be dimensionless. The difference CGS extensions differ by the powers of the speed of light.

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i am given that my constant wavenumber K=l₀/2A.where lo is the constant speed and A=4*pi*m*e^2/ℏ²n₀^1/3 .If i take A to have same units as 1/v^2 then dimensions of K are not satisfied.what to do now??? –  zara Sep 27 '12 at 7:54
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Multiply it by whatever power of the speed of light $c$ is needed to fix the dimensions. –  Luboš Motl Sep 27 '12 at 9:47
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