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Angular momentum is given by: $n\frac{h}{2\pi}$ where $n=1,2,3...$

And angular momentum is also given by: $\sqrt{l(l+1)}\frac{h}{2\pi}$

By the first equation angular momentum should be $\frac{h}{2\pi}$ but by the second it should be $0$.

And if you say that in first equation $n$ is also $0$ then first radius of hydrogen should be $0\pu{A^\circ}$ instead of $0.53\pu{A^\circ}$.

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You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular momentum is thus given by

$$ l_n=m_e|\vec{r}_n\times\vec{v}_n|, $$ where $m_e$ the electron mass. Bohr postulated that the angular momentum must be quantized (this is the semi-classical part) resulting in a set of orbitals.

Though Bohr's model works remarkably well it is not perfect because electrons are not really particles in the classical sense, but also behave like waves. A better treatment (but not the best) of the hydrogen atom is therefore to use quantum mechanics. In quantum mechanics the electron is described by a wavefunction that is spread out over all space and has a certain shape. For $s$ orbitals, such as the lowest orbital in the hydrogen atom, the wavefunction is spherical symmetric and has no angular momentum. The second expression you mention is the quantum mechanical operator to obtain the angular momentum.

To conclude, you cannot compare the angular momentum of the Bohr atom to the Schrodinger atom because they are based on different assumptions. Note further that for principal quantum number $n$, $\ell=0, 1, \ldots, n-1$.

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