de-Broglie wavelength,λ=mvh​ where h= Planck's constantm=mass of the particlev=velocity of the particle(a) It is given that the speed of an electron and proton are equal.It is clear from the above equation thatλ∝m1​ As mass of a proton, mp​>mass of an electron, me​, the proton will have a smaller wavelength compared to the electron.(b)λ=ph​(∵p=mv)So, when the proton and the electron have same momentum, they will have the same wavelength.(c) de-Broglie wavelength (λ) is also given byλ=2mE​h​ where E=energy of the particle.Let the energy of the proton and electron be E . Wavelength of the proton,λp​=2mp​E​h​…(1) Wavelength of the electron,λe​=2me​E​h​…(2 Dividing (2) by (1), we get:​λp​λe​​=mp​​me​​​⇒λp​λe​​<1⇒λe​<λp​​ It is clear that the proton will have smaller wavelength compared to the electron.