Question

De broglie wavelength of electron accelerated by potential v is lambda what is de broglie wavelenght of electron which is accelerated by 4v potential

Answer 1

de-Broglie wavelength for an electron is given by lambda=h/√2meV. So, if you plugin the values of h(Planck's constant),m(mass of electron),e(elementary charge of electron), in the above equation you get a value, lambda=12.27A°/√V. So, according to your problem, the answer is 1.227A°, where 1A°=10^(-10) m.