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(D) The de-Broglie wavelength $(\lambda)$ of a particle is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the

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$\lambda \propto E^{-\frac{1}{2}}$

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(D) The de-Broglie wavelength $(\lambda)$ of a particle is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the particle.We know that the relationship between kinetic energy $(E)$ and momentum $(p)$ is $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.Substituting this into the de-Broglie wavelength equation,we get $\lambda = \frac{h}{\sqrt{2mE}}$.Since $h$ and $m$ (mass of the particle) are constants,we can conclude that $\lambda \propto \frac{1}{\sqrt{E}}$,which is equivalent to $\lambda \propto E^{-\frac{1}{2}}$.

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(E)$ as

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