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Q.
The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
ManipalManipal 1980Dual Nature of Radiation and Matter
Solution:
de Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
$\lambda=\frac{h}{\sqrt{2 m k_{B} T}}$
where $m$ is the mass of the neutron
$k_{B}$ is the Boltzmann constant
$h$ is the Planck's constant
Here, $m =1.67 \times 10^{-27} k g$
$ k_{B}=1.38 \times 10^{-23} J K^{-1} $
$h=6.63 \times 10^{34} J s$
$\therefore \lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T}} $
$= \frac{3.08 \times 10^{-34} \times 10^{25}}{\sqrt{T}} m $
$= \frac{30.8 \times 10^{-10}}{\sqrt{T}}=\frac{30.8}{\sqrt{T}}\,\mathring{A}$