1. Use the formula for orbital angular momentum

For an electron in an orbital with azimuthal quantum number $l$, the magnitude of orbital angular momentum is

$L = \sqrt{l(l+1)}\,\frac{h}{2\pi}$

2. For the $2s$ orbital

For an $s$-orbital,

$l=0$

So,

$L = \sqrt{0(0+1)}\,\frac{h}{2\pi} = 0$

Thus, for $2s$ orbital, orbital angular momentum is

$0$

3. For the $2p$ orbital

For a $p$-orbital,

$l=1$

So,

$L = \sqrt{1(1+1)}\,\frac{h}{2\pi} = \sqrt{2}\,\frac{h}{2\pi}$

Thus, for $2p$ orbital, orbital angular momentum is

$\sqrt{2}\,\frac{h}{2\pi}$

4. Match with the options

So the values for electrons in $2s$ and $2p$ orbitals, respectively, are:

$0 \text{ and } \sqrt{2}\,\frac{h}{2\pi}$

This matches Option B.

5. Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

Hence, they agree.