Can the equation $E = h\nu$ be used not only for light, but for radio waves? (Since they are all part of the electromagnetic spectrum) Can the equation $E = h\nu$  (or $E = hf$) be used to find the energy of the particles from the electromagnetic spectrum? I mean, for example light is part of the EM, a radio wave is also part of the EM, can we find the energy of the particles that make up the radio waves? Light is made up of photons, and with the equation we can find their energy, well, can we find the energy of the particles that radio waves (or other parts of the EM, except the visible light) are made of? Sorry for the stupid question, I am just getting into Quantum Mechanics (Khan Academy).
 A: Yes you can. The relation $E=h\nu$ is quite general and lies in the heart of quantum mechanics. It gives the energy of any radiation that is vibrating at a frequency of $\nu$.
Although later, we generally replace this with $E=\hbar \omega$, because $\omega$ forms a nice 4-vector with $\textbf{k}$, which is the wave number. But it also basically says the same thing since $\hbar = h/2\pi$ and $\omega=2\pi\nu$.
So, I would suggest you to get familiar with this notation. But you can use the formula you mention as long as you are not dealing with Relativity :)
A: To add to the existing answers: indeed you can ─ but the resulting energies are tiny, and you need to look to physical phenomena with a correspondingly tiny energy scale to observe single-photon effects. 
To put some numbers into it, the principal transition in hydrogen, 1s-2p, has a transition energy of 10eV and a wavelength of 120 nm. If you go to the radio domain, say, 1 MHz, then the corresponding energy at $E=h\nu$ is in the pico-electronvolt regime, which is much smaller than any useful transition in matter.
There are, of course, some very useful transitions that fall in the radio-wave regime, most notably the hydrogen 21cm line at a frequency of ~1.4 GHz and a photon energy of about 6 µeV, which is an absolutely tiny spacing caused by the extremely weak interaction between the magnetic dipole moments of the proton and the electron in what's known as hyperfine structure (a correction on the fine structure of the spectrum, which is itself a correction over the main spectral series). Since that interaction is weak, the 21cm emission is also very weak, and you need specialized equipment to measure it in the lab, or astronomical amounts of hydrogen if you want to see this emission 'in the wild'. (Luckily, of course, the universe does have astronomical amounts of hydrogen lying around for astronomers to study.)
More generally, there are relatively few relevant single-photon phenomena in that energy range, and you need to that kind of detailed spacing at µeV (or smaller) energy resolutions if you want to see radio-wave-regime photon energies.
A: Yes, yes, of course it can be used. This experimentally derived property is valid for all EM waves. 
A: Just adding one important detail. You wrote (v3):

Light is made up of photons, and with the equation we can find their energy, well, can we find the energy of the particles that radio waves (or other parts of the EM, except the visible light) are made of?

Radio waves are also made up of photons! All electromagnetic radiation consists of photons, and for all of them $E = h\nu$ holds as the others have already said.
To reiterate, there is no fundamental difference between "different kinds" of EM radiation. What we call light is only distinguished because our visual receptors interact only with light of those frequencies.
