Answer
(a) $ \sqrt{2} \;\hbar $
(b) $ \sqrt{6} \;\hbar $
Work Step by Step
This problem asks for the minimum and maximum values of
$$
\sqrt{L_x^2 + L_y^2}
$$
where $ L_x $ and $ L_y $ are the components of the angular momentum along the $x$ and $y$ axes, respectively.
The total angular momentum $ L $ is given by:
$$
L = \sqrt{l(l+1)} \hbar
$$
where $ l = 2 $:
$$
L = \sqrt{2(2+1)} \hbar = \sqrt{6}\; \hbar\tag 1
$$
We know that:
$$
L^2 = L_x^2 + L_y^2 + L_z^2
$$
Thus:
$$
L_x^2 + L_y^2 = L^2 - L_z^2
$$
Plug from (1);
$$
L_x^2 + L_y^2 = 6\hbar^2 - L_z^2 \tag 2
$$
The possible values of $ L_z $ are given by:
$$
L_z = m_l \hbar
$$
$$\color{blue}{\bf [a]}$$
The minimum value of $ \sqrt{L_x^2 + L_y^2} $, for the minimum value, $ L_z $ is at its maximum, $ L_z = 2\hbar $
Plug into (2);
$$
L_x^2 + L_y^2 = (\sqrt{6} \hbar)^2 - (2 \hbar)^2 = 6\hbar^2 - 4\hbar^2 = 2\hbar^2
$$
So,
$$
\boxed{\sqrt{L_x^2 + L_y^2} = \sqrt{2} \;\hbar\;}
$$
$$\color{blue}{\bf [b]}$$
The maximum value of $ \sqrt{L_x^2 + L_y^2} $, for the maximum value, $ L_z = 0 $,
$$
L_x^2 + L_y^2 = (\sqrt{6} \hbar)^2 - 0^2 = 6\hbar^2
$$
So,
$$
\boxed{\sqrt{L_x^2 + L_y^2} = \sqrt{6}\; \hbar}
$$
