A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of V esu. If e and m are charge and mass of the electron respectively, then the value of h//lamda (where lamda is the wavelength associated with the electron wave) is given by

A stream of electrons from a heated filament was passed between two ch

1.	A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of V esu. If e and m are charge and mass of the electron respectively, then the value of h//lamda (where lamda is the wavelength associated with the electron wave) is given by
Answer» `me V` `2 me V` `SQRT(me V)` `sqrt(2 meV)` Solution :`LAMDA = (h)/(p) or (h)/(lamda) = p = MV` `KE = (1)/(2)mv^(2)` so that `v = sqrt((2KE)/(m)) " But " KE = eV` `:. v = sqrt((2EV)/(m)) :. (h)/(lamda) = mv = m sqrt((2eV)/(m)) = sqrt(2meV)`