# Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use u_(avg) = sqrt((8k_BT)/(pi m))

Jul 18, 2017

I got $\text{0.0794 nm}$, or $\text{79.4 pm}$, over twice its atomic radius.

Well, assuming that the $\text{He}$ atoms are all moving in the same direction, and follow a Maxwell-Boltzmann distribution:

$f \left(u\right) = 4 \pi {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{3 / 2} {u}^{2} {e}^{- m {u}^{2} / 2 {k}_{B} T}$

they thus have the average speed

${u}_{a v g} = {\int}_{0}^{\infty} u f \left(u\right) \mathrm{du} = \left[. . .\right] = \sqrt{\frac{8 {k}_{B} T}{\pi m}}$,

where:

• ${k}_{B} = 1.38065 \times {10}^{- 23}$ is the Boltzmann constant in $\text{J/K}$, or $\text{kg"cdot"m"^2"/s"^2cdot"K}$.
• $T$ is the temperature in $\text{K}$.
• $m$ is the per-particle mass in $\text{kg}$, i.e. $M / {N}_{A} = m$, where $M$ is the molar mass in $\text{kg/mol}$ and ${N}_{A} = 6.0221413 \times {10}^{23} {\text{mol}}^{- 1}$.

We then use this average speed as the velocity $v$ in the de Broglie relation

$\lambda = \frac{h}{m v}$,

where $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$, or $\text{kg"cdot"m"^2"/s}$, is Planck's constant and $m$ is as defined before,

since the particles are assumed to all be moving in the same direction so that the velocity is the speed.

Therefore, the wavelength is:

$\textcolor{b l u e}{\lambda} = \frac{h}{m} \times \sqrt{\frac{\pi m}{8 {k}_{B} T}}$

$= \sqrt{\frac{\pi {h}^{2}}{8 m {k}_{B} T}}$

$= \sqrt{\left(\pi \left(6.626 \times {10}^{- 34} \cancel{\text{kg")cdot"m"^cancel(2)"/"cancel("s"))^2)/(8(0.0040026 cancel("kg")"/"cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23)))(1.38065 xx 10^(-23) cancel("kg")cdotcancel("m"^2)"/"cancel("s"^2)cdotcancel"K")(298.15 cancel("K}}\right)\right)}$

$= 7.938 \times {10}^{- 11}$ $\text{m}$

$=$ $\textcolor{b l u e}{\text{0.0794 nm}}$,

which is in the gamma-ray region of the EM spectrum.