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Question:

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

 

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

Answer:

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

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

Planck’s constant, h = 6.6 × 10 −34 Js

Kinetic energy (K) and velocity (v) are related as:

K = 1/2 mnv2                                    ... (1)

De Broglie wavelength (λ) and velocity (v) are related as:

λ = h/ mnv                                       ....(2)

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

Hence, the kinetic energy of the neutron is 6.75 × 10 −21 J or 4.219 × 10 −2 eV.

 

(b) Temperature of the neutron, T = 300 K

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

Average kinetic energy of the neutron:

K' = 3/2 kT

= 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J

The relation for the de Broglie wavelength is given as:

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


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