Question

Neutrons in thermal equilibrium with matter at $27^{\circ} C$ can be thought to behave like an ideal gas. Assuming them to have a speed of $v_{\text {rms }}$, their de-Broglie wavelength is $\lambda$ (in nm). Find $\left(\frac{156}{11}\right) \lambda$.
(Take $m_{n}=1.69 \times 10^{-27} kg, \left.k=1.44 \times 10^{-23} J / K , h=6.60 \times 10^{-34} Js \right)$

Answer 1

$v_{ rms }=\sqrt{\frac{3 k T}{m_{n}}} ; \lambda=\frac{h}{p}$
$ \Rightarrow \lambda=\frac{h}{m_{n} \times v_{ rms }}=\frac{h}{\sqrt{3 k T m_{n}}}$
$\Rightarrow \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 1.44 \times 10^{-23} \times 1.69 \times 10^{-27} \times 300}}$
$=\frac{2.2 \times 10^{-10}}{1.2 \times 1.3}$
$\Rightarrow \frac{156}{11} \cdot \lambda=\frac{156 \times 2.2 \times 10^{-10}}{11 \times 1.2 \times 1.3}$
$=\frac{220 \times 10^{-10}}{11}$
$=\frac{22}{11} nm =2 \,nm$