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

The correct option is Option 4 (the graph in Image 3 representing a straight line passing through the origin).

To find the correct graph showing the variation of $\frac{1}{{\lambda }^2}$ with kinetic energy $E$, we can derive the mathematical relationship between these variables step-by-step:

1. The de Broglie wavelength $\lambda$ of a free particle of momentum $p$ is given by the relation:
$\lambda =\frac{h}{p}$
where $h$ is Planck's constant.

2. The kinetic energy $E$ of a particle of mass $m$ is related to its momentum $p$ by the equation:
$E=\frac{p^2}{2m}$
Rearranging this equation to express momentum $p$ in terms of kinetic energy $E$ gives:
$p=\sqrt{2mE}$

3. Substituting this expression for $p$ back into the de Broglie wavelength formula yields:
$\lambda =\frac{h}{\sqrt{2mE}}$

4. Squaring both sides of the equation to eliminate the square root gives:
${\lambda }^2=\frac{h^2}{2mE}$

5. Taking the reciprocal of both sides leads to the final relation:
$\frac{1}{{\lambda }^2}=\left(\frac{2m}{h^2}\right)E$

6. Since the mass $m$ of the particle and Planck's constant $h$ are constants, the quantity $\frac{2m}{h^2}$ is also a constant. Therefore, we can write:
$\frac{1}{{\lambda }^2}\propto E$

This is a linear equation of the form $y=kx$, where $y=\frac{1}{{\lambda }^2}$, $x=E$, and the slope $k=\frac{2m}{h^2}$ is positive. The graph corresponding to this relation is a straight line starting from the origin $O$ and extending with a constant positive slope, as depicted in the graph of Image 3 (Option 4).