A particle of charge `q` mass `m` and energy `E` has de-Broglie wave length `lambda` .For a particle of charge `2q` mass `2m` and energy `2E` the de-Broglie wavelength is

The de Broglie wavelength lambda of a particle

A particle moving with kinetic energy E_(1) has de Broglie wavelength lambda_(1) . Another particle of same mass having kinetic energy E_(2) has de Broglie wavelength lambda_(2). lambda_(2)= 3 lambda_(2) . Then E_(2) - E_(1) is equal to

The de Broglie wavelength associated with particle is

The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as

The de-Broglie wavelength lambda associated with charged particle of charge q, mass m and potential difference V is

For particles having same K.E., the de-Broglie wavelength is

A free particle with initial kinetic energy E, de-Broglie wavelength lambda , enters a region where in it has a potential V, what is the new de-Broglie wavelength?