CUET · PHYSICS · PYQ PAPER 2023
An electron is accelerated through a potential difference of 144 V. The de-broglie wavelength associated would be approximately:
- A 10 nm
- B 1 nm
- C 0.1 nm
- D 0.01 nm
Answer & Solution
Correct Answer
(C) 0.1 nm
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{1.226}{\sqrt{V}} \text{ nm} \) \( \lambda = \frac{1.226}{\sqrt{144}} \) \( \lambda = \frac{1.226}{12} \approx 0.102 \text{ nm} \)
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