For electron in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are:

For any electron in an atomic orbital, the magnitude of the orbital angular momentum $$L$$ is given by
$$L = \sqrt{l(l+1)}\,\hbar$$ where $$l$$ is the azimuthal (orbital) quantum number and $$\hbar = \dfrac{h}{2\pi}$$.

Case 1: Electron in the $$2s$$ orbital
For an $$s$$-orbital, $$l = 0$$.
Therefore,
$$L_{2s} = \sqrt{0(0+1)}\,\hbar = 0$$.

Case 2: Electron in the $$2p$$ orbital
For a $$p$$-orbital, $$l = 1$$.
Therefore,
$$L_{2p} = \sqrt{1(1+1)}\,\hbar = \sqrt{2}\,\hbar = \sqrt{2}\,\dfrac{h}{2\pi}$$.

So the orbital angular momenta are 0 for $$2s$$ and $$\sqrt{2}\,\dfrac{h}{2\pi}$$ for $$2p$$.

Hence, the correct option is Option D.