Step 1: Understanding the Concept
The de Broglie wavelength is \(\lambda=\dfrac{h}{p}=\dfrac{h}{mv}\).
Step 2: Speed after falling
For a particle dropped from rest, \(v^2=2gy\), so \(v=\sqrt{2gy}\).
Step 3: Calculation
\(\lambda=\dfrac{h}{m\sqrt{2gy}}\). Here \(h\), \(m\) and \(g\) are constants, so \(\lambda\propto y^{-1/2}\).
Final Answer:
The wavelength varies as y to the power minus one half, which is option (D).
\[ \boxed{\text{(D) }y^{-1/2}} \]