Typically, in quantum mechanics, bound states are quantized and free/scattering states are not. This is because bound states, by the mere fact that they're constrained to a certain area, will have to satisfy certain boundary conditions, and these conditions won't be able to be satisfied in a continuous range.
The classic example of this is the infinite square well potential, where $V(x) = 0$ if $0<x<a$, and $V(x) =\infty$ elsewhere. Then, the particle will have zero probability of appearing outside of the well, and will have to satisfy the zero-potential Schrödinger equation $E\psi = -\frac{\hbar^{2}}{2m}\nabla^{2}\psi$ inside of the well. For simplicity, we'll only consider one-dimensional motion.
In this case, we see right away that the basis states to our solutions have to satisfy $\psi = A\sin\left(\frac{\sqrt{2mE}}{\hbar}x+\phi\right)$, and we also know that the wave function must be continuous, and that it is restricted to be zero for $x<0$ and $x>a$. We can satisfy the first boundary condition by choosing $\phi=0$, but the second one is not satisfied for all values of the energy. Instead, it is necessary that $\frac{\sqrt{2mE}}{\hbar}a=n\pi$, where $n$ is some integer. Thus, the allowed energies of 'pure' states of this system are quantized, and take the values $E_{n} = \frac{n^{2}\pi^{2}\hbar}{2m}$.
For any other bound state, you will find yourself using similar logic about boundary conditions, albiet with much, much more complexity. Note that, however, it is also the case that we can construct a general state out of the energy eigenstates $\Psi = \sum a_{n}\psi_{n}$, and that the expectation value for the energy of $\Psi$ will be $\sum|a_{n}|^{2}E_{n}$, so the values for the "average" value of a state are still allowed to be continuous (and in the case of the infinite square well, can actually take any value greater than the ground state energy).