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CUET · PHYSICS · PYQ PAPER 2023

An electron is accelerated through a potential difference of 144 V. The de-broglie wavelength associated would be approximately:

  1. A 10 nm
  2. B 1 nm
  3. C 0.1 nm
  4. D 0.01 nm
Verified Solution

Answer & Solution

Correct Answer

(C) 0.1 nm

Step-by-step Solution

Detailed explanation

\( \lambda = \frac{1.226}{\sqrt{V}} \text{ nm} \) \( \lambda = \frac{1.226}{\sqrt{144}} \) \( \lambda = \frac{1.226}{12} \approx 0.102 \text{ nm} \)