Confusion regarding the orbital angular momentum of mesons in an excited state I came across the statement that Mesons can have a spin of a whole number (e.g. 1 or 2) instead of the regular 0, as a result of the orbital angular momentum they have in their excited states, with this orbital angular momentum always being a whole number in units of $\hbar$. 
But the way I remember the formula for orbital angular momentum, namely $L = \sqrt {l(l+1)}\hbar$, does not deliver whole numbers.
Can somebody clear up the confusion?
Thank you
 A: You do not give a link for "came accross" which would give the context.
Here is a table of mesons as determined in experiments, with their quantum numbers.
Here is a similar list where the quark content is displayed.
They are made up of a quark and an antiquark (makeup column in the second link) pair. Quarks and antiquarks have spins , but the interaction between them is the strong interaction which does not lead to orbitals as with electrons around nuclei.( See this to understand the complexity of imaging hadrons in terms of quarks and antiquarks and gluons ).
Angular momentum  conservation allows to see the spin of a meson as zero or one, from the addition of  a quark antiquark pair. For higher spins in the mesons one needs to go to lattice QCD which is fairly successful in modeling strong interactions, here is a talk.

After discussion in comments:
$\ell$ is the quantum number, and integer,  associated with the angular momentum. This is an integer number associated with the  quantization of the angular momentum of a given energy state .
$$L^2 = \ell(\ell+1)\hbar^2$$
So it is the square of the angular momentum that is quantized and $\ell$ is the quantum number characterizing it. It is similar to n, being the principle quantum number characterizing the quantized energy levels.
The simplicity of the relation between $L$ and $\ell$ allows to talk of conservation of angular momentum in terms of adding spin quantum numbers with angular momentum quantum numbers invoking conservation of angular momentum, since the algebraic connection is simple  (something that is not true for the principle quantum number and energy).  
A: 
the formula for orbital angular momentum, namely $L = \sqrt {l(l+1)}\hbar$

The property encapsulated by this formula is the fact that if $\mathbf J$ is an angular momentum operator, and the system is in a state with well-defined total angular momentum $J^2$, then that value will be
$$
J^2 = j(j+1)\hbar^2
$$
for $j$ a nonnegative integer or a half-integer. When we say that a given system "has spin X", we are referring to the quantum number $j$, not to the value of $J^2$. That use of language is not referring to the eigenvalue of $J^2$ (which is indeed equal to $j(j+1)\hbar^2$ nor to the square root of that eigenvalue, $\sqrt{j(j+1)}\hbar$: the latter generally isn't an integer multiple of $\hbar$, but it is not what we refer to in the usage you're concerned about.
