Question:
Two identical parallel metal plates A and B, having fine holes at their centers are connected to a power supply as shown in the figure. An electron having energy 200 eV is directed to pass through these holes from A to B. The de Broglie wavelength of the electron when it comes out of plate B is:
Two identical parallel metal plates A and B, having fine holes at their centers are connected to a power supply as shown in the figure. An electron having energy 200 eV is directed to pass through these holes from A to B. The de Broglie wavelength of the electron when it comes out of plate B is:
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Wavelength $\lambda$ is inversely proportional to square root of Voltage!
Updated On: Jun 6, 2026
- 0.713 Å
- 2.012 Å
- 1.754 Å
- 1.227 Å
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The Correct Option is A
Solution and Explanation
Step 1: Concept
de Broglie wavelength $\lambda = \frac{12.27}{\sqrt{V}} \text{Å}$.
Step 2: Meaning
Final energy $E = 200 \text{ eV} + 100 \text{ eV} = 300 \text{ eV}$.
Step 3: Analysis
$\lambda = \frac{12.27}{\sqrt{300}} \text{Å} = \frac{12.27}{17.32} \approx 0.71 \text{Å}$.
Step 4: Conclusion
The wavelength is approximately 0.713 Å.
Final Answer: (A)
de Broglie wavelength $\lambda = \frac{12.27}{\sqrt{V}} \text{Å}$.
Step 2: Meaning
Final energy $E = 200 \text{ eV} + 100 \text{ eV} = 300 \text{ eV}$.
Step 3: Analysis
$\lambda = \frac{12.27}{\sqrt{300}} \text{Å} = \frac{12.27}{17.32} \approx 0.71 \text{Å}$.
Step 4: Conclusion
The wavelength is approximately 0.713 Å.
Final Answer: (A)
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