Calculate the

(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

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v

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

h = 6.6 × 10^-34Js

p = 4.04 × 10^-24 Kgms^-1

So putting the values in equation we get

So the de Broglie wavelength of electron is 0.164 nm

Note: we could have also used the direct relation when an electron is accelerated through a potential difference of V volts its de Broglie wavelength is given by

1A^o = 10^-10m