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, 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:

… (1)

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

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 m² s⁻² K⁻¹

Average kinetic energy of the neutron:

The relation for the de Broglie wavelength is given as:

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