Question

Find the ratio of de-Broglie wavelength associated with a proton and an \( \alpha \)-particle having the same kinetic energy.

Hint:

For particles with the same kinetic energy, de-Broglie wavelength is inversely proportional to the square root of mass.

Answer 1

The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mK}} \] For equal kinetic energy: \[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}} \] Since the mass of an alpha particle is four times that of a proton: \[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = 2 \] \[ \boxed{2} \]