Answer:
Thermal de Broglie Wavelength is 4.11×10-7 m
Explanation:
The thermal wavelength for a perfect quantum gas in any number of measurements and for a summed up connection between vitality and energy.
Thermal de Broglie Wavelength is generally the normal de Broglie wavelength of the gas particles in a perfect gas at the predetermined temperature
to find the thermal de Broglie Wavelength
we have the formula
Λ = h / (2∙π∙m∙K∙T)^1/2
Putting values in the formula where h is Planck constant, m is mass of the particles which is taken as an average mass per atom of rubidium, k is Boltzmann constant, and T is the thermodynamic temperature that is given above in the question.
Λ = (6.63×10⁻34 J∙s) / (2π(1.42×10⁻25 kg)(1.38×10⁻²³ J∙K⁻¹)(
0.20uK))
= 4.11×10-7