Question

An electron of mass m when accelerated through a potential difference v, has de broglie wavelength λ. The de-broglie wavelength associated with a particle of mass m/4 and charge e accelerated through the same potential difference is nλ. Then find the value of n

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.