Question

The de-Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is

• A. $\frac{3.08}{\sqrt{T}}\mathring{A}$
• B. $\frac{0.308}{\sqrt{T}}\mathring{A}$
• C. $\frac{0.0308}{\sqrt{T}}\mathring{A}$
• D. $\frac{30.8}{\sqrt{T}}\mathring{A}$

Answer 1

Answer: (D)

de Broglie wavelength of neutrons in thermal equilibrium at temperature $T$ is
$\lambda =\frac{h}{\sqrt{2m{k}_{B}T}}$
where $m$ is the mass of the neutron
$k_B$ is the Boltzmann constant
$h$ is the Planck's constant
Here, $m=1.67\times 10^{-27}kg$
$k_B=1.38\times 10^{-23}J{K}^{-1}$
$h=6.63\times 10^{34}Js$
$\therefore \lambda =\frac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.38\times 10^{-23}\times T}}$
$=\frac{3.08\times 10^{-34}\times 10^{25}}{\sqrt{T}}m$
$=\frac{30.8\times 10^{-10}}{\sqrt{T}}=\frac{30.8}{\sqrt{T}}\mathring{A}$