CUET · PHYSICS · PYQ PAPER 2025
A proton has double the kinetic energy of an \( \alpha \)- particle. The ratio of the de-Broglie wavelength of the proton to that of \( \alpha \)- particle will be
- A \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{\sqrt{2}}{1}\)
- D \(\frac{2}{1}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{2}}{1}\)
Step-by-step Solution
Detailed explanation
\( \lambda = \frac{h}{\sqrt{2mK}} \) \( \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha K_\alpha}{m_p K_p}} \) \( \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{(4m_p) K_\alpha}{m_p (2K_\alpha)}} \) \( \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4}{2}} = \sqrt{2} \)
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