PhysicsMedium145×since 2002Q6588The kinetic energy of an electron, α\alphaα-particle and a proton are given as 4 K,2 K4 \mathrm{~K}, 2 \mathrm{~K}4 K,2 K and K\mathrm{K}K respectively. The de-Broglie wavelength associated with electron (λe),α(\lambda \mathrm{e}), \alpha(λe),α-particle ((λα)((\lambda \alpha)((λα) and the proton (λp)(\lambda p)(λp) are as follows:Aλα<λp<λe\lambda \alpha<\lambda p<\lambda eλα<λp<λeBλα>λp>λe\lambda \alpha>\lambda p>\lambda eλα>λp>λeCλα=λp<λe\lambda \alpha=\lambda p<\lambda eλα=λp<λeDλα=λp>λe\lambda \alpha=\lambda p>\lambda eλα=λp>λeCheck answerSkip