$$\lambda = \frac{h}{p}$$
where:
* $\lambda$ is the de-Broglie wavelength in meters
* $h$ is the Planck constant ($6.626 \times 10^{-34}$ J s)
* $p$ is the momentum of the atom in kg m/s
The momentum of an atom can be calculated using the following formula:
$$p = mv$$
where:
* $m$ is the mass of the atom in kg
* $v$ is the velocity of the atom in m/s
The velocity of an atom at absolute temperature T K can be calculated using the following formula:
$$v = \sqrt{\frac{3kT}{m}}$$
where:
* $k$ is the Boltzmann constant ($1.381 \times 10^{-23}$ J/K)
* $T$ is the absolute temperature in Kelvin
* $m$ is the mass of the atom in kg
Substituting the expressions for $p$ and $v$ into the formula for the de-Broglie wavelength, we get:
$$\lambda = \frac{h}{\sqrt{3mkT}}$$
This is the de-Broglie wavelength of an atom at absolute temperature T K.
