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 1

(a) De Broglie wavelength 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
Kinetic energy (K) and velocity (v) are related as:

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 m² s⁻² K⁻¹ 

Average kinetic energy of the neutron:

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