Q 12-11-008NEETNEET 2024Top questionEasy
The graph which shows the variation of $\left(\dfrac{1}{\lambda^2}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle) :
Answer: (B) see figure
$\lambda = \dfrac{h}{\sqrt{2mE}}$, so
$$\frac{1}{\lambda^2} = \frac{2m}{h^2}E$$
$\dfrac{1}{\lambda^2}$ is directly proportional to $E$: a straight line through the origin, which is option (2).
Solution by Sreeraj P, M.Sc Physics