The graph which shows the variation of (1/λ²)  and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle) :

The correct option showing the variation of 1/λ² with kinetic energy E is:

Step 1: Understand the Relationship between de Broglie Wavelength and Kinetic Energy
According to de Broglie's hypothesis, the wavelength λ associated with a free particle of mass m and momentum p is given by:

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

where h is Planck's constant.

The kinetic energy E of a particle of mass m is related to its momentum p by:

$E=\frac{p^2}{2m}\Rightarrow p=\sqrt{2mE}$

Step 2: Express 1/λ² in terms of E
Substituting the value of p into de Broglie's wavelength equation:

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

Squaring both sides of the equation gives:

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

Taking the reciprocal of both sides:

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

Step 3: Analyze the Graph Characteristics
Since m and h are constant, the term $\left(\frac{2m}{h^2}\right)$ is a positive constant C.
Thus, the equation is of the form:

$y=mx$

where:
• $y=\frac{1}{{\lambda }^2}$ is plotted on the vertical y-axis.
• $x=E$ is plotted on the horizontal x-axis.
• The slope is $\frac{2m}{h^2}>0$.

This represents a straight line passing through the origin O with a positive slope.
Therefore, the graph showing a straight line originating from the origin represents the correct variation of 1/λ² with E.