CUET · PHYSICS · PYQ PAPER 2025
An electron is accelerated through a potential difference of \( V = 36 \text{ V} \). The de-Broglie wavelength associated with it will be
- A \( 2.05 \times 10^{-10} \text{ m} \)
- B \( 2.26 \times 10^{-9} \text{ m} \)
- C \( 1.26 \times 10^{-10} \text{ m} \)
- D \( 2 \times 10^{-7} \text{ m} \)
Answer & Solution
Correct Answer
(A) \( 2.05 \times 10^{-10} \text{ m} \)
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{h}{\sqrt{2me V}} \) \( \lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{\sqrt{2 \times 9.109 \times 10^{-31} \text{ kg} \times 1.602 \times 10^{-19} \text{ C} \times 36 \text{ V}}} \) \( \lambda = 2.05 \times 10^{-10} \text{ m} \)
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