Energy of elementary particles and Planck constant The energy of a photon is proportional to Planck's constant. Is this the case for the energy of other elementary particles?
 A: This is a question that in the beginning of experimentally exploring quantum effects was an obvious one and was proposed by de Broglie to be similar :
$E=ν*h$
Thus a frequency $ν$ can be defined for each particle, which has given rise to the term matter wave.
Caution:
It is  important to note  that  the so called "matter wave" is a confusing term. An individual particle's  energy $E$ is not spread out in space, as is very clearly seen in experiments with particles one at a time, in the same link:

The interference pattern appears with an accumulation  of electrons diffracted through the same set up. The individual electrons leave a point on the screen, characteristic of a particle. The same is true with single photons at a time as seen here . .
Quantum mechanics is a probabilistic theory, not a deterministic one, one can only compute the probability of detecting a particle at (x,y,z) and all elementary particles are point particles.
A: You are reading the formula in the "wrong" way in terms of physics. The energy is proportional to the frequency. Plank constant is just the proportionality constant which depends on the system of units. We can take it to be one for example. And then it is obvious that the relationship is between energy and frequency.
The proportionality is a relationship between variables. It means that when one variable changes, the other one does too, and in a specific way. "Proportionality" between a variable and a constant does not make sense.
A: For an elementary particle, its energy levels $E$ are the eigenvalues of the Time Independent Schrödinger Equation:
$$\hat{H}\Psi=E\Psi$$
where $\hat{H}$ is the Hamiltonian (total energy) operator:
$$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf{r})$$
with $\hbar=\frac{h}{2\pi}$.
It follows that these eigenvalues $E$ will always contain $h^2$:
$$E\propto h^2$$
As examples look up the energies for the $\text{1D}$ particle in an infinite box and the hydrogen atom.
For the $\text{P1DB}$:
$$E_n=\frac{n^2h^2}{8mL^2}$$
For $n=1,2,3,...$
For the Quantum Harmonic Oscillator:
$$E_n=(n+\frac12)\hbar \omega$$
For $n=1,2,3,...$
For the hydrogen atom:
$$E_n=-\frac{me^4}{32 \pi^2 \epsilon_0 \hbar^2 n^2}$$
For $n=1,2,3,...$
So in each of these cases the energy levels $E_n$ depend on the quantum number $n$ and $h$ or $\hbar$.
