NEETPhysicsDual Nature of Radiation and MatterMCQ+4 / −1
The graph which shows the variation of and its kinetic energy, is (where is de Broglie wavelength of a free particle):
- A
- B
- C
- D
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Correct answer: D
de-Broglie wavelength $$\lambda=\frac{h}{P}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$$ where $E=\frac{1}{2} m v^2$
Squaring both sides,
$$\begin{aligned} & \lambda^2=\frac{h^2}{4 m^2 E} \\ & \Rightarrow \frac{1}{\lambda^2}=\text { (constant) } E \end{aligned}$$
Graph passes through origin with constant slope.
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