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

The Correct Option is D

de Broglie wavelength of neutrons in thermal equilibrium at temperature T is $?= \frac {h}{\sqrt {2mk_BT}} $ 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} J \, s $ $\therefore ?= \frac {6.63 \times 10^{-34}}{\sqrt {2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T}}$ $= \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} $