The graph which shows the variation of $(\frac{1}{\lambda^2})$ and its kinetic energy, $E$ is (where $\lambda $ is de Broglie wavelength of a free particle):

The correct answer is option (4) :

de-Broglie wavelength $\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$ where $E=\frac{1}{2}mv^2$

Squaring both sides,

$\lambda^2 =\frac{h^2}{4m^2E}$

$⇒\frac{1}{\lambda^2}= (constant)E$

Graph passes through origin with constant slope.