Question:
The ratio of the respective de Broglie wavelengths of two particles with kinetic energy of 0.02 eV and 2 eV is ________.
The ratio of the respective de Broglie wavelengths of two particles with kinetic energy of 0.02 eV and 2 eV is ________.
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Higher kinetic energy means a shorter de Broglie wavelength.
Updated On: Jun 26, 2026
- 1:1
- 10:1
- 1:10
- $1:\sqrt{10}$
- $\sqrt{10}:1$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$, where $K$ is kinetic energy.
Step 2: Meaning
For the same particle (mass $m$), $\lambda \propto \frac{1}{\sqrt{K}}$.
Step 3: Analysis
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{2}{0.02}} = \sqrt{100}$.
Step 4: Conclusion
$\frac{\lambda_1}{\lambda_2} = 10$, which is a ratio of 10:1. Final Answer: (B)
The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$, where $K$ is kinetic energy.
Step 2: Meaning
For the same particle (mass $m$), $\lambda \propto \frac{1}{\sqrt{K}}$.
Step 3: Analysis
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{2}{0.02}} = \sqrt{100}$.
Step 4: Conclusion
$\frac{\lambda_1}{\lambda_2} = 10$, which is a ratio of 10:1. Final Answer: (B)
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