An electron of mass m has de-Broglie wavelength λ when accelerated through potential difference V . When proton of mass M , is accelerated through potential difference 9 V , the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)

Step-by-step solution

When electron or any charged particle is accelerated through potential difference V, then kinetic energy gained is given by E = e V ...... (i) E = 1 2 m v 0 2 = p 2 2 m = h 2 2 m . λ 2 ...... (ii) ∴     e V = h 2 2 m . λ 2   ⇒ λ = h 2 m e V ....... (iii) When proton of mas M is accelerated through a potential difference of 9 V , then the de - Broglie wavelength obtained is λ ′ = h 2 M   e 9 V = h 3 × 2 M e V × m m ∴   λ ′ = &#