When a potential drop of V volt is applied on a stationary neutron. then de broglie wavelength-

An electron of mass m has de broglie wavelength lamda 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 ) .

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 ) .

An alpha-"particle" is accelerated through a potential difference of V volts from rest. The de-Broglie's wavelengths associated with it is.

An alpha-"particle" is accelerated through a potential difference of V volts from rest. The de-Broglie's wavelengths associated with it is.

An alpha-"particle" is accelerated through a potential difference of V volts from rest. The de-Broglie's wavelengths associated with it is.

Show that the de Broglie wavelength of charged particles accelerated through a potential of V volt is given by lamda = h//sqrt(2mqV) . Hence show that the de-Broglie wavelength of electron is (12.27)/(sqrtV)Å .

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 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 :-