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Assuming ideal nature of gases in accordance to kinetic theory of gases,

The expression of equipartition of energy which assigns $\frac{k_BT}{2}$ energy ($k_B$= boltzmann constant and $T$ = absolute temperature) to each degree of freedom ($f$) and gives total kinetic energy of a molecule as $\frac{fk_BT}{2}$

Can we equate this to general expression of kinetic energy ?

$\frac{fk_BT}{2} = \frac{mv^2}{2}$

which gives $v= \sqrt\frac{fkT}{m}$ where $m$ = mass of molecule.

If it's true then what kind of velocity is it? And can we substitute this velocity in expression of de broglie wavelength?

$\lambda= \frac{h}{mv}$ ($h$ = Planck's constant)

But if it's true, then it will contradict the expression which comes by taking root mean square velocity

$v = \sqrt\frac{3RT}{m}$ (R = universal gas constant)

and gives $\lambda= \frac{h}{\sqrt{3mkT}}$ for all molecules independent of their degree of freedom.

mindfulamatter
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Amarnath Parasar
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1 Answers1

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De-Broglie's expression for the wavelength is $\lambda = \frac{h}{|\vec{p}|}$, where $\vec{p} = m\vec{v}$ is the linear momentum of the particle. As such, any ideal gas particle will always have only 3 translational degrees of freedom (DOF's) in $x,y,z$, so that $\frac{mv^2}{2} = \frac{3k_BT}{2}$. Further DOF's related to rotation and vibration are internal to the molecule, and are used during the equipartition of total energy $H$. Refer to this related post. These might be helpful as well: Equipartition of Energy, also look at the expression for the total energy $H$ in this PhySE post.

mindfulamatter
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