A particle is dropped from a height ' y '. The de-Broglie...

When a particle is dropped and falls a distance $y$, its initial velocity is zero. The velocity $v$ of the particle after falling a distance $y$ is given by the equation of motion:

$v^2 = u^2 + 2gy$

$v = \sqrt{2gy}$

The momentum of the particle is $p = mv = m\sqrt{2gy}$.

The de-Broglie wavelength is given by $\lambda = \dfrac{h}{p}$.

Substituting the value of momentum, we get:

$\lambda = \dfrac{h}{m\sqrt{2gy}}$

Therefore, $\lambda \propto y^{-1/2}$.

Answer: $y^{-1/2}$