Zero Angular momentum at a Non-Zero distance in Quantum Mechanics In Bohr atomic theory We learn that only those orbits are possible for which angular momentum is intger multiple of $h$.ie.
$$mvr=\frac{nh}{2\pi}$$
Here n is used for angular quantum number,That can take value $n=1,2,3..$.
Now in quantum theory we learn that for Hydrogen atom the wave function for ground state where $n=1$ ,$l=0$ and $m=0$. The wave function given by 
$$\psi_{100}=Ce^{-r/a_0}$$ where $C$ is some constant. Here $l$ is angular quantum number.The probability for a particle to be at distance $r$ to $r+dr$ is given by apart from some constant 
$$P(r)=|\psi_{100}|^2r^2dr=Cr^2e^{-2r/a_0}$$ That suggest that particle is most probable to found around $a_0$. So How it is possible that particle is at non-zero distance from the orgin(not doing a radial motion) but have a zero angular momentum? How bohr's theory differes from this result?    
 A: For a particle to have zero angular momentum it means that it's in an eigenstate of operator 
$$ {\bf L}^2 = ({\bf r} \times {\bf p})^2 = -\hbar^2 ({\bf r}\times\nabla)^2 $$
with eigenvalue 0. That is 
$$ ({\bf r}\times\nabla)^2 \psi({\bf r}) = 0$$
Solving this equation in radial cooridantes leads to a solution
$$ \psi(r,\theta,\phi) = f(r)$$
where $f$ is an arbitrary function (smooth enough).
In conclusion, a particle whose wavefunction is spherically symetric will have zero angular momentum. Regerdless of its expected value of distance from the center of the system of coordinates.
A: It is easy: just recall the classical-mechanics definition of angular momentum:
$$ \bf L = \bf r \times p $$
If $\bf p$ is along the direction of $\bf r$, $\bf L$ is certainly zero, regardless if the magnitudes of both quantities are not zero.
A: Even in classical mechanics, a particle can have zero angular momentum at a non-zero distance from the origin if its velocity is purely radial. You have to remeber that the angular momentum norm is $||\overrightarrow{L}|| = ||r||\, ||p|| \, |\sin(\theta)|$, where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{p}$.
In quantum mechanics, the picture is slightly different, because the particle does not have a definite position nor a definite momentum, and is instead described by a wavefunction $\psi(\overrightarrow{r})$ such that the probability density for the particle to be at $\overrightarrow{r}$ is $|\psi(\overrightarrow{r})|^2$. What about $\overrightarrow{p}$? This is were the correspondance with classical mechanics breaks down, as $\overrightarrow{p}$ cannot be determined independently of $\overrightarrow{r}$. Instead, it can be shown from the postulates of quantum mechanics that $\overrightarrow{p}$ should be replaced by $-i \hbar \overrightarrow{\nabla}$ when we intergrate it against something over $\overrightarrow{r}$.
This allows to calculate the mean angular momentum of any state described by a wavefunction $\psi$:
$$\left\langle \overrightarrow{L} \right\rangle = \int \psi^*(\overrightarrow{r}) (\overrightarrow{r} \times \overrightarrow{p})\psi(\overrightarrow{r})\overrightarrow{dr} = - i \hbar \int \psi^*(\overrightarrow{r}) \overrightarrow{r} \times (\overrightarrow{\nabla}\psi(\overrightarrow{r}))\overrightarrow{dr}$$
But in the case of a rotationally symmetric wavefunction $\psi(\overrightarrow{r})$ (such as any wavefunction with $l = 0$ in the Bohr theory), $\overrightarrow{\nabla}\psi(\overrightarrow{r})$ is directed along $\overrightarrow{r}$, so that the cross-product in the integral is $0$.
This was of course to be expected: $\left\langle \overrightarrow{L} \right\rangle$ is a vector, but since the wavefunction describing the state is rotationally symetric, any "choice" of direction for $\left\langle \overrightarrow{L} \right\rangle$ would be completely arbitrary, so it must be that $\left\langle \overrightarrow{L} \right\rangle = \overrightarrow{0}$.
A: The wave function can be interpreted as describing an oscillating movement instead of an orbital one. But it is odd that the probability of finding an electron at $r = 0$ is zero.
It only suggests that it is frustrating to carry the intuition of our macroscopic world to the quantum level. It is better to learn how to calculate.

A: There have already been satisfactory answers to this question above. I would like to add a sort of "intuitive" explanation.
In classical mechanics, when you have rotational symmetry in a problem (for example, the earth solely rotating about the sun), it essentially means that you could perform a rotation about the center and obtain another valid solution, namely, another plane of rotation. But at any given time, the particle is only in one of these orbits and thus it might be difficult to think of orbits with 0 angular momenta throughout.
Coming to quantum mechanics, the picture is different. You could think of the spherically symmetric wave functions as being a superposition of all the classical planar orbits in three-dimensional space. And in essence, as a result of the superposition, you can think of the particle whizzing about in random directions and resulting in the expectation value of angular momentum turning out to be 0, at a non-zero distance from the atom.
It's a hand-wavy argument but I believe it helps in clarifying some questions above.
