Calculate the

(a) momentum, and


(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

(a) When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by


K = Vq, which imparts it with velocity


We know kinetic energy of a Particle with mass m kg and velocity v is given by



Equating both equations



We get velocity of the particle as



But we know momentum is given by relation


P = mv


Where P is the momentum of particle having mass m and moving with velocity v


So we get



Here Potential difference


V = 56 volts


q = 1.6 × 10-19 C


m = 9.1 × 10-31Kg


so momentum of the electron,




So momentum of electron is 4.04 × 10-24 Kgms-1


(b) Now we know de Broglie wavelength of a Particle is given by relation


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