Q. 17 A4.0( 11 Votes )

# For what kinetic

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

Or we can say momentum of particle is

P = h/ 𝜆

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

So substituting we get

or

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

so putting P = h/ 𝜆

we get kinetic energy of particle as

Where h is the Planck’s constant, m is mass of particle having de Broglie wavelength 𝜆

Here,

The de Broglie Wavelength is, 𝜆 = 1.40 × 10–10m

The particle is neutron and mass of neutron is, m = 1.66 × 10–27Kg

putting these values we get the kinetic energy of neutron as

So kinetic energy of the neutron is 6.75 × 10-21J

Converting it to eV

We know

1 eV = 1.6 × 10-19J

i.e.

1J = (1/1.6 × 10-19)eV

So we get kinetic energy of neutron as

Or we get Kinetic energy of neutron as 0.042 Ev

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