A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as (a) λp = λn > λe > λα (b) λα < λp = λn < λe (c) λe < λp = λn > λα (d) λe = λp = λn = λα
Correct option is B) Kinetic energy of particle, K=21mv2ormv=2mK de Broglie wavelength, λ=mvh=2mKh For the given value of K,λαm1 ∴λp:λn:λe:λα=mp1:mn1:me1:mα1 Since mp=mn,henceλp=λn As mα>mp,thereforeλalpha<λp As me<mn,thereforeλn<λn Hence λalpha<λp=λn<λe
