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):

The de-Broglie wavelength is given by:
$$\lambda =\frac{h}{\sqrt{2mE}}$$
Squaring both sides:
$${\lambda }^2=\frac{h^2}{2mE}\Rightarrow \frac{1}{{\lambda }^2}=\left(\frac{2m}{h^2}\right)E$$
This equation is of the form $y=mx$, which represents a straight line passing through the origin.Thus, the graph of $\frac{1}{{\lambda }^2}$ vs $E$ is a straight line.