CUET · PHYSICS · PYQ PAPER 2023
A proton and \(\alpha\)-particle have same kinetic energy. The de-Broglie wavelength ratio \( \frac{\lambda_p}{\lambda_{\alpha}} \) is equal to:
- A \( \frac{m_{\alpha}}{m_p} \)
- B \( m_{\alpha} \times m_p \)
- C \( \sqrt{\frac{m_{\alpha}}{m_p}} \)
- D \( \sqrt{\frac{m_p}{m_{\alpha}}} \)
Answer & Solution
Correct Answer
(C) \( \sqrt{\frac{m_{\alpha}}{m_p}} \)
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{h}{\sqrt{2mK}} \) \( \frac{\lambda_p}{\lambda_{\alpha}} = \frac{h/\sqrt{2m_p K}}{h/\sqrt{2m_{\alpha} K}} \) \( \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}} \)
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