To solve this problem, we need to understand the relationship between the de Broglie wavelength $\lambda$ and kinetic energy $E. \\$The de Broglie wavelength is given by:$\\ \lambda = \frac{h}{p} \\$where $h$ is Planck's constant and $p$ is the momentum of the particle.$\\$For a particle with mass $m$ and velocity $v,$ the momentum $p$ is:$\\ p = mv \\$The kinetic energy $E$ of the particle is:$\\ E = \frac{1}{2}mv^2 \\$Solving for $v$ in terms of $E: \\ v = \sqrt{\frac{2E}{m}} \\$Substitute $v$ into the momentum equation:$\\ p = m \cdot \sqrt{\frac{2E}{m}} = \sqrt{2mE} \\$Substitute $p$ into the de Broglie wavelength equation:$\\ \lambda = \frac{h}{\sqrt{2mE}} \\$Now, find $\frac{1}{\lambda^2}: \\ \frac{1}{\lambda^2} = \left(\frac{\sqrt{2mE}}{h}\right)^2 = \frac{2mE}{h^2} \\$This shows that $\frac{1}{\lambda^2}$ is directly proportional to $E. \\$Therefore, the graph of $\frac{1}{\lambda^2}$ versus $E$ is a straight line passing through the origin.$\\$The correct graph is Option 2.