Atomic StructureMedium
Question
The graph which shows the variation of $\left( \frac{1}{\lambda^{2}} \right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):
Options
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Solution
Sol. [Key Concept: Graph of momentum vs kinetic energy (p = √(2mE))] de-Broglie wavelength $\lambda = \frac{h}{P} = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}$ where $E = \frac{1}{2}mv^{2}$
Squaring both sides,
$$\begin{matrix} & \lambda^{2} = \frac{h^{2}}{4m^{2}E} \\ & \ \Rightarrow \frac{1}{\lambda^{2}} = (\text{~}\text{constant}\text{~})E \end{matrix}$$
Graph passes through origin with constant slope.
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