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22 March, 10:47

Rubidium atoms are cooled to 0.20 μK in an atom trap.

What is their de Broglie wavelength?

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  1. 22 March, 10:59
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    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
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