\(\sqrt2 :1\)

Let the length of the rectangle be \( l \) and breadth be \( b \).
The radius of the semicircle, half the length of the rectangle, and the breadth form a right-angled triangle.
Applying the Pythagorean theorem:
\[ \left(\frac{l}{2}\right)^2 + b^2 = 2^2 \]
The area of the rectangle is \( l \times b \).
To maximize the area, we can use the AM-GM inequality or set the two squared terms equal:
\[ \left(\frac{l}{2}\right)^2 = b^2 \]
This yields:
\[ \frac{l}{2} = b \Rightarrow l = 2b \]
The ratio of the length to the breadth is therefore:
\[ \frac{l}{b} = \frac{2}{1} \]
The correct option is (D): \( 2:1 \)