Question

Calculate the wavelength of de-Broglie waves associated with a neutron at room temperature of ${27}^{\circ }\text{C}$. Given the mass of the neutron, $m=1.67\times {10}^{-27}\text{kg}$, Boltzmann constant, $k=1.38\times {10}^{-23}{\text{J K}}^{-1}$ and $h=6.6\times {10}^{-34}\text{Js}$.

• A. $2.9\mathring{A}$
• B. $1\mathring{A}$
• C. $1.4\mathring{A}$
• D. $5\mathring{A}$

Answer 1

The correct option is C $1.4\mathring{A}$

Given,
$T={27}^{\circ }\text{C}=300\text{K};m=1.67\times {10}^{-27}\text{kg}$

$k=1.38\times {10}^{-23}{\text{J K}}^{-1};h=6.6\times {10}^{-34}\text{Js}$

The kinetic energy of a thermal neutron is,
$KE=\frac{3}{2}kT$

$\frac{p^2}{2m}=\frac{3}{2}kT$

Thus, $p=mv=\sqrt{3mkT}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{3mkT}}$

$=\frac{6.6\times {10}^{-34}}{\sqrt{3\times 1.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}$

$=\frac{6.6\times {10}^{-34}}{4.55\times {10}^{-24}}$

$\lambda =1.45\times {10}^{-10}\text{m}=1.45\mathring{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.