An electron of charge `e` and mass `m` is accelerated from rest by a potential difference `V`. the de Broglie wavelength is

A particle of charge q and mas m si accelerated from set through a potential difference V . Its de Broglie wavelength is equal to

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 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 is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 xx 10^(-2) nm, the potential difference is:

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 xx 10^(-2) nm, the potential difference is:

The mass of an electron is m, its charge e and it is accelerated from rest through a potential difference V. The Kinetic energy of the electron in joules will be:

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 when accelerated through a potential difference V, has de Broglie wavelength lambda . The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be :-

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