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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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