CUET · PHYSICS · PYQ PAPER 2025
A proton accelerated through a potential difference of V volts has a de-Broglie wavelength \( \lambda \) associated with it. In order to get the same wavelength with an \( \alpha \)-particle, the required accelerating potential is
- A V/16
- B V/8
- C 4V
- D 8V
Answer & Solution
Correct Answer
(B) V/8
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{h}{\sqrt{2mqV}} \) \( \lambda_p = \frac{h}{\sqrt{2m_p e V}} \) \( \lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e) V_\alpha}} = \frac{h}{\sqrt{16m_p e V_\alpha}} \) \( \frac{h}{\sqrt{2m_p e V}} = \frac{h}{\sqrt{16m_p e V_\alpha}} \)…
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