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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

  1. A V/16
  2. B V/8
  3. C 4V
  4. D 8V
Verified Solution

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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