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(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of t

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(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is its kinetic energy.Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mE}$.Rearranging the terms,we find $\frac{1}{\lambda^2} = \left(\frac{2m}{h^2}\right) E$.This equation is of the form $y = mx$,where $y = \frac{1}{\lambda^2}$,$x = E$,and the slope $m = \frac{2m}{h^2}$ is a constant.Therefore,the graph of $\frac{1}{\lambda^2}$ versus $E$ is a straight line passing through the origin.

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

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