A particle is dropped from height H. The de-Broglie wavelength of the particle as a function of height is proportional to:

$H^{-1/2}$

The correct answer is Option (4) → $H^{-1/2}$

The De-Broglie Wavelength is -

$λ=\frac{h}{momentum (P)}$

and,

Momentum, $P=mv$

$=m\sqrt{2g(H-h)}$  $[v=\sqrt{2g(H-h)}]$

Hence,

$λ=\frac{h}{m\sqrt{2g(H-h)}}⇒λ∝\frac{1}{\sqrt{H}}$