A proton has double the kinetic energy of an $α$-particle. The ratio of the de-Broglie wavelength of the proton to that of $α$-particle will be

$\frac{\sqrt{2}}{1}$

The correct answer is Option (3) → $\frac{\sqrt{2}}{1}$

de-Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$

For proton:

$\lambda_p = \frac{h}{\sqrt{2m_p K_p}}$

For $\alpha$-particle:

$\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} K_{\alpha}}}$

Given: $K_p = 2K_{\alpha}$

Ratio:

$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}K_{\alpha}}{m_p K_p}}$

$= \sqrt{\frac{m_{\alpha}K_{\alpha}}{m_p (2K_{\alpha})}}$

$= \sqrt{\frac{m_{\alpha}}{2m_p}}$

Mass of $\alpha$-particle = $4m_p$

$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{2m_p}} = \sqrt{2}$

Answer: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{2}$