Question

The de broglie wavelength of neutrons in thermal equlibrium at temperature $T$ is

• A. $\frac{3.08}{\sqrt{T}}\stackrel{0}{A}$
• B. $\frac{0.308}{\sqrt{T}}\stackrel{0}{A}$
• C. $\frac{0.0308}{\sqrt{T}}\stackrel{0}{A}$
• D. $\frac{30.8}{\sqrt{T}}\stackrel{0}{A}$

Answer 1

The correct option is D $\frac{30.8}{\sqrt{T}}\stackrel{0}{A}$

de broglie wavelenght of neutrons in thermal equlibrium at temperature $T$ is
$\lambda =\frac{h}{\sqrt{2m{K}_{B}T}}$
Here, $m=1.67\times {10}^{-27}kg{K}_B=1.38\times {10}^{-23}J{K}^{-1}\left(Boltzmannconstant\right)h=6.63\times {10}^{-34}Js$

$\therefore \lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 1.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times T}}=\frac{30.8\times {10}^{-10}}{\sqrt{T}}=\frac{30.8}{\sqrt{T}}\stackrel{0}{A}$