CUET · PHYSICS · PYQ PAPER 2025
An electron accelerated with potential V has a de-Broglie wavelength \( \lambda \) associated with it. If the potential is changed to 4 V, the wavelength becomes
- A \( \lambda/4 \)
- B \( \lambda/2 \)
- C \( \lambda \)
- D \( 2\lambda \)
Answer & Solution
Correct Answer
(B) \( \lambda/2 \)
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{h}{\sqrt{2meV}} \) \( \lambda' = \frac{h}{\sqrt{2me(4V)}} \) \( \lambda' = \frac{h}{\sqrt{4 \cdot 2meV}} \) \( \lambda' = \frac{1}{2} \frac{h}{\sqrt{2meV}} \) \( \lambda' = \frac{\lambda}{2} \)
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