Answer :

Here we are given kinetic energy of electron


K = 120 eV


We know 1 eV = 1.6 × 10-19 J


i.e. kinetic Energy ,K = 120 × 1.6 × 10-19 J = 1.92 × 10-17J


(a) We know kinetic energy of a particle is given by the relation



Where K is the kinetic energy of a particle of mass m moving with speed v


Now multiplying L.H.S. and R.H.S. of the equation by m we get



or


But we know momentum is given by relation, p = mv


So, substituting we get



or


i.e. p = √2mK


Or we can say momentum P of any particle can be expressed in terms of its mass m and kinetic energy K as


p = √2mK


Particle is electron so we have mass of electron


m = 9.1 × 10-31Kg


kinetic energy of particle


K = 1.92 × 10-17J


Putting the values in equation we get



= 5.91 × 10-24 Kgms-1


So we get momentum of electron is 5.91 × 10-24 Kgms-1


(b) But we know momentum is given by relation


P = mv


Where P is momentum of particle of mass m moving with speed v


So we get speed of electron is


v = P/m


here the momentum of electron


P = 5.91 × 10-24 Kgms-1


Mass of electron


m = 9.1 × 10-31Kg


so putting these values we get the speed of electron as



or we can say speed of electron is 6.5 × 106 ms-1


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


𝜆 = h/mv


Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v and h is Planck’s constant


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 substituting we get de Broglie wavelength of Particle as


𝜆 = h/P


value of Planck’s constant


h = 6.6 × 10-34Js


and momentum of electron is


P = 5.91 × 10-24 Kgms-1


Putting these values we get



= 0.112 nm


So the wavelength of the electron is 0.112 nm


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