Question 11.17:
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x 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 300K.

Hint: \(K=\frac{p^{2}}{2m}\)

Step 1: Find the momentum of a neutron.
According to De Broglie
Step 2: Find the kinetic energy of a neutron.
Hence, the kinetic energy of the neutron is 6.75 x 10-21 J or 4.219 x 10-2 eV.

Hint: \(\lambda^{\prime}=\frac{h}{\sqrt{2 K m }}\)

Step 1: Find the average kinetic energy of the neutron.
The average kinetic energy of the neutron:

Step 2: Find the wavelength of a neutron.
The relation for the de Broglie wavelength is given as λ=h2Km
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.