What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

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