Is E=3/2 Kb T or E=Kb T?

[dinky]

Energy and frequency E=hv

I have a simple question that is that is E=3/2x K_b x T, where k_b is Boltzmann constant.
I understand that this is involve 3 degree of freedom, but as i was reading through the forum, i come across 1 equation stating E=k_b x T. So is it true? and what application would this apply to?

And with regards to quantum physics, since kinetic energy= 1/2MV², and =3/2k_bT,
yet, λ=h/p=h/mv, where h is Planck constand and p is momentum, and λ is wavelength.

Since λ=V/F, where v is velocity and F is frequency,
then this would mean that V/F=h/mv, and I would get hxF=mV²,
and since E=hxv, where E is energy, h is Planck constant, and v is velocity,
would that mean that E=mV² which I don't understand as I thought E was suppose to be 1/2mv², yet now the 1/2 has disappear.

Thanks a lot for the help, a bit confused with all the energy, frequency and wavelength.

 

[DrClaude]

I have a simple question that is that is E=3/2x K_b x T, where k_b is Boltzmann constant.
I understand that this is involve 3 degree of freedom, but as i was reading through the forum, i come across 1 equation stating E=k_b x T. So is it true? and what application would this apply to?

The equipartition tells us that
$$
\langle E \rangle = \frac{f}{2} k_\mathrm{B} T
$$
i.e., each quadratic degree of freedom contributes $k_\mathrm{B} T/2$ to the average energy.

For kinetic energy only, since $E \propto p_x^2 +p_y^2 + p_z^2$, there are $f=3$ quadratic degrees of freedom, from which we get the first equation in the OP.

The second equation is of course obtained for $f=2$, which would be the case for, e.g., a 2D monatomic ideal gas or a harmonic oscillator (since $E$ is then obtained from the sum of a term in $p^2$ and a term in $x^2$). The vibration of diatomic molecules is most often modeled as a harmonic oscillator.

And with regards to quantum physics, since kinetic energy= 1/2MV², and =3/2k_bT,
yet, λ=h/p=h/mv, where h is Planck constand and p is momentum, and λ is wavelength.

Since λ=V/F, where v is velocity and F is frequency,

The latter $\lambda$ is the wavelength of a classical wave propagating at a velocity $v$, and shouldn't be confused with the former $\lambda$, which is the de Broglie wavelength.