The de-Broglie wavelength associated with a charged particle is given by λ=ph where, h= planck's constant and p= momentum =2mKE (here, KE is the kinetic energy of the charged particle) ⇒λ=2mKEh For proton and α-particle, the wavelengths are respectively given as, λp=2 mpKEph and λα=2 mαKEαh∴λαλp=2 mpKEP2 mαKEα Here, KEα=KEp and mα=4 mp Substituting these above mentioned relations in Eq. (i), we get ⇒λαλP=mp4 mP=12 or λP:λα=2:1