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Question -

(a) For what kinetic energy ofa neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?

(b) Also find the deBroglie wavelength of a neutron, in thermal equilibrium with matter, having anaverage kinetic energy of (3/2) kT at 300 K.



Answer -

(a) De Brogliewavelength of the neutron, λ = 1.40 × 10−10 m

Mass of a neutron, mn =1.66 × 10−27 kg

Planck’s constant, h = 6.6× 10−34 Js

Kineticenergy (K) and velocity (v) are related as:

 … (1)

DeBroglie wavelength (λ) and velocity (v) are related as:

Using equation (2) inequation (1), we get:

Hence, the kinetic energy of the neutron is6.75 × 10−21 J or 4.219 × 10−2 eV.

(b) Temperature of theneutron, T = 300 K

Boltzmann constant, k =1.38 × 10−23 kg m2 s−2 K−1

Average kinetic energy ofthe neutron:

The relation for the deBroglie wavelength is given as:

Therefore,the de Broglie wavelength of the neutron is 0.146 nm.

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