Rubidium atoms are cooled to 0.20 μk in an atom trap. what is their de broglie wavelength?

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

that would be because of both gravity and friction. so     c

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west