An electron of charge e and mass m is acceleration from rest with a potential V. What is the de Broglie wavelength of the electron?

An electron (charge = e, mass = m) is accelerated from rest through a potential difference of V volt. What will be the de Broglie wavelength of the electron?

An electron of mass 'm' is accelerated by a potential difference V and the corresponding de-Broglie wavelength is lamda The de Broglie wavelength of a proton of mass M if it is accelerated by the same potential difference is.

An electron (mass m) and a proton (mass M) are accelerated with same potential difference, then

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. Its final speed will be

An alpha -particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelengths associated with them.

if rahe radius of the first Bohr orbit is x, then find the de Broglie wavelength of electron in third orbit.

If the radius of first Bohr orbit of H atom is x then de Broglie wavelength of electron in 3rd orbit is nearly

An electron is accelerated by a photential difference V and as a result, its de Broglie wavelength beocmes lamda . If the applied potential difference was 2V, the de Broglie wavelength would have been

For a moving electron, the de Broglie wavelength is 1 Å . What will be the de Broglie wavelength if (i) the momentum of electron be doubled, (ii) the kinetic energy of the electron be doubled?

If the speed of an electron is v and the speed of another particle is 3v, ratio of the de Broglie wavelength of the electron to that of the particle is (10^4)/(8.13) . Is the particle a proton or an alpha -particle? Justify your answer.