MCQ
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):
- A
- B
- ✓
- D
✓
Answer
Correct option: C.
c
de-Broglie wavelength $\lambda=\frac{h}{P}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$ where $E=\frac{1}{2} m v^2$
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,
$\lambda^2=\frac{h^2}{4 m^2 E}$
$\Rightarrow \frac{1}{\lambda^2}=\text { (constant) } E$
Graph passes through origin with constant slope.
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