Question:
If the kinetic energy of a free electron doubles, it�s de-Broglie wavelength changes by the factor
If the kinetic energy of a free electron doubles, it�s de-Broglie wavelength changes by the factor
Updated On: Apr 26, 2024
- $2$
- $\frac{1}{2}$
- $\sqrt{2}$
- $\frac{1}{\sqrt{2}}$
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The Correct Option is D
Solution and Explanation
de-Broglie wavelength,
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2.m.\left(K.E\right)}}$
$\therefore \lambda\,\propto \frac{1}{\sqrt{K.E}}$
If K.E is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2.m.\left(K.E\right)}}$
$\therefore \lambda\,\propto \frac{1}{\sqrt{K.E}}$
If K.E is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$
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