Q. 124.3( 14 Votes )

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 ร 10-19 C

m = 9.1 ร 10-31Kg

so momentum of the electron,

So momentum of electron is 4.04 ร 10-24 Kgms-1

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

๐ = h/mv

Where ๐ 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

๐ = h/P

h = 6.6 ร 10-34Js

p = 4.04 ร 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

1Ao = 10-10m

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