The de Broglie wavelength (lambda) depends on | vedclass

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(C) The kinetic energy $E$ of a particle of mass $m$ and momentum $p$ is given by $E = \frac{p^2}{2m}$.From this,the momentum is $p = \sqrt{2mE}$.The

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$m^{-1/2}E^{-1/2}$

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(C) The kinetic energy $E$ of a particle of mass $m$ and momentum $p$ is given by $E = \frac{p^2}{2m}$.From this,the momentum is $p = \sqrt{2mE}$.The de Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.Substituting the value of $p$,we get $\lambda = \frac{h}{\sqrt{2mE}}$.This can be rewritten as $\lambda = \frac{h}{\sqrt{2}} m^{-1/2} E^{-1/2}$.Therefore,$\lambda$ is proportional to $m^{-1/2} E^{-1/2}$.

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

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