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.

To find the correct graph showing the variation of $\frac{1}{{\lambda }^2}$ with kinetic energy $E$ for a free particle, we can use the de Broglie wavelength formula:

$\lambda =\frac{h}{p}$

The relationship between momentum $p$ and kinetic energy $E$ for a particle of mass $m$ is given by:

$p=\sqrt{2mE}$

Substituting this momentum expression into the de Broglie wavelength formula, we get:

$\lambda =\frac{h}{\sqrt{2mE}}$

Squaring both sides of the equation:

${\lambda }^2=\frac{h^2}{2mE}$

Taking the reciprocal of both sides gives:

$\frac{1}{{\lambda }^2}=\left(\frac{2m}{h^2}\right)E$

Since the mass $m$ of the particle and Planck's constant $h$ are constants, the term in parentheses is constant. Therefore:

$\frac{1}{{\lambda }^2}\propto E$

This proportional relationship represents a linear equation of the form $y=kx$ passing through the origin, where $y=\frac{1}{{\lambda }^2}$ and $x=E$. The graph is a straight line starting at the origin (O) with a positive, constant slope.

Comparing the options:
- image_0.png shows a curve bending upwards (concave up).
- image_1.png shows a curve decreasing as E increases.
- image_2.png shows a curve bending downwards (concave down).
- image_3.png shows a straight line passing through the origin.

Thus, the straight line in image_3.png is the correct graph, corresponding to Option 4.