Q. 32 A4.6( 5 Votes )

# Obtain the de Bro

Answer :

Given:

Kinetic energy of neutron, E = 150 eV = 150 × 1.6 × 10^{-19}

E = 2.4 × 10^{-17} J

Mass of neutron, m_{n} = 1.675 × 10^{-27} Kg

Kinetic energy of a particle is given by,

We can write,

mv = (2 × m × KE)^{0.5}

P = (2 × m × KE)^{0.5} ...(1)

Where,

m = mass of electron

v = velocity of electron

p = momentum of particle

By putting the equation (1) in de- Broglie equation, we get,

…(2)

Now substituting the values in eq (2) we get,

λ = 2.327 × 10^{-12}m

A neutron cannot be used in diffraction experiment as lattice spaces are of the order of few 10^{-10} m, whereas the wavelength of a neutron beam is 100 times smaller.

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