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
Concept. $\lambda = \dfrac{h}{\sqrt{2mE}}$.
Step 1. Relation. $\dfrac{1}{\lambda^2} = \dfrac{2m}{h^2}E$: a straight line through the origin, graph (2).
Answer. Graph (2) (Option B).
Solution by Sreeraj P, M.Sc Physics