Neglecting the variation of mass with velocit | vedclass

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

Vedclass · Accepted Answer

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

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(C) The de Broglie wavelength $\lambda$ of a particle is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the particle.The kinetic energy $E$ of a particle of mass $m$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.Substituting the expression for $p$ into the de Broglie wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.Since $h$ and $m$ are constants,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$,which can be written as $\lambda \propto E^{-1/2}$.Therefore,the correct option is $C$.

Neglecting the variation of mass with velocity,the wavelength associated with an electron having a kinetic energy $E$ is proportional to

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