Calculate the (a) momentum, and (b) the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
(a) Potential difference, V=56 volts At equilibrium, Kinetic energy is accelerating potential. 2mv2=eV v=4.44×106m/s P=mv ⇒9.1×10−31×4.44×106=4.04×10−24 where m is the mass of the electron. (b) De Broglie wavelength is given by λ=V12.27×10−10 ⇒5612.27×10−10=0.164 nm
