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
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \frac{\sqrt{2}}{1} \)
The de Broglie wavelength is \( \lambda = h/p \), and a particle of mass \( m \) and kinetic energy \( K \) has momentum \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \;\Rightarrow\; \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha K_\alpha}{m_p K_p}} \)
\( m_\alpha = 4m_p \)
\( K_p = 2K_\alpha \)
\( \displaystyle \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p \times K_\alpha}{m_p \times 2K_\alpha}} = \sqrt{2} = \frac{\sqrt{2}}{1} \)