Question

An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

Answer 1

De Broglie wavelength of particle of mass m in terms of kinetic energy $$(E_k)$$ is $$\lambda=\dfrac{h}{\sqrt{2mE_k}} \propto \dfrac{1}{m}$$ for the same kinetic energy.
(i) Out of given particles, the mass of alpha particle is maximum, so de Broglie wavelength associated with alpha particle is the shortest.
(ii) As mass of electron is least, so electron has largest de-Broglie wavelength.