# (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be

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.

Solution:

(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:

$K=\frac{1}{2} m_{n} v^{2} \ldots$ (1)

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

$\lambda=\frac{h}{m_{n} v}$   ...(2)

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

$K=\frac{1}{2} \frac{m_{n} h^{2}}{\lambda^{2} m_{n}{ }^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{n}}$

$=\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2 \times\left(1.40 \times 10^{-10}\right)^{2} \times 1.66 \times 10^{-27}}=6.75 \times 10^{-21} \mathrm{~J}$

$=\frac{6.75 \times 10^{-21}}{1.6 \times 10^{-19}}=4.219 \times 10^{-2} \mathrm{eV}$

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:

$K^{\prime}=\frac{3}{2} k T$

$=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300=6.21 \times 10^{-21} \mathrm{~J}$

The relation for the de Broglie wavelength is given as:

$\lambda^{\prime}=\frac{h}{\sqrt{2 K^{\prime} m_{n}}}$

Where,

$m_{n}=1.66 \times 10^{-27} \mathrm{~kg}$

$h=6.6 \times 10^{-34} \mathrm{~J}_{\mathrm{S}}$

$K^{\prime}=6.75 \times 10^{-21} \mathrm{~J}$

$\therefore \lambda^{\prime}=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27}}}=1.46 \times 10^{-10} \mathrm{~m}=0.146 \mathrm{~nm}$

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