Q. 164.1( 11 Votes )

# An electron and a

Answer :

(a) In order to find momentum, we know de Broglie wavelength for any particle is given by

𝜆 = h/P

Or P = h/𝜆

Where h is Planck’s constant

h = 6.63 × 10^{-34} Js

and P is the momentum of particle

here we are given the de Broglie wavelength

𝜆 = 1.00 nm = 1.00 × 10^{-9} m

__Note:____we can see that momentum only depends on de Broglie wavelength and Planck’s Constant which is same for both electron or Photon(Light Particle) so the momentum for both will also be same__

Putting values we get momentum

So momentum of electron and photon is 6.63 × 10^{-25}Kgms^{-1}

(b) we know energy of an photon is given by the relation

E = hc/𝜆

Where E is the energy of Photon having wavelength 𝜆 and speed c,h is Planck’s constant

h = 6.63 × 10^{-34} Js

Speed of light or photon in free space

c = 3 × 10^{8} m/s

And we are given wavelength

𝜆 = 1.00 nm = 1.00 × 10^{-9} m

Putting these values we get

Converting it to eV

We know

1 eV = 1.6 × 10^{-19}J

i.e.

1J = (1/1.6 × 10^{-19})eV

So we get energy of Photon as

so we get energy of Photon as 1.24KeV

(c) 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 velocity 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

Where m is mass of Particle moving with velocity v

So substituting we get

or we get kinetic energy of particle K in terms of momentum P and mass of particle m as

The particle is electron and mass of electron is

m = 9.1 × 10^{-31}Kg

Here the momentum of electron

P = 6.63 × 10^{-25}Kgms^{-1}

Putting these values in above equation we have

Converting it to eV

We know

1 eV = 1.6 × 10^{-19}J

i.e.

1J = (1/1.6 × 10^{-19})eV

So we get Kinetic energy of electron as

so we get Kinetic energy of electron as 1.51eV

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