If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is

Step-by-step solution

Given, _ e = _ p Since, de-Broglie wavelength, aligned &= h m v = h p = h 2 m K &= h 2 m K aligned where, K is kinetic energy. array ll _ e _ p & = m_ p m_ e K_ p K_ e _ e _ e & = m_ p m_ e K_ p K_ e [From Eq. (i)] array K_ e K_ p = m_ p m_ e Squaring both sides, we get array ll & K_ e K_ p = m_ p m_ e since & m_ p >m_ e & K_ e >K_ p array