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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):
Detailed Solution
de Broglie wavelength $\lambda = \frac{h}{\sqrt{2mE}}$
$\Rightarrow \frac{1}{\lambda^2} = \frac{2mE}{h^2}$, i.e. $\frac{1}{\lambda^2} \propto E$
So the graph of $\frac{1}{\lambda^2}$ vs E is a straight line passing through the origin.