If de-broglie wavelength of an electron of mass m when accelerated through a potential difference V is λ. Then de-Broglie wavelength associated with a proton of mas M accelerated through the same potential difference will be

Let the proton and electron is accelerated with the same potential V, then de Broglie wavelength of the electron will be given as,λ=h2meV ; where h is Planck's constant.The de Broglie wavelength of the proton will be,λp=h2MeVSo the ratio of both wavelengths will be,λλp=Mm⇒λp=λmM