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

Given:

Temperature T = 0.20 μK

To determine:

The de Broglie wavelength of Rubidium atoms

Explanation:

The de broglie wavelength (λ) is related to the temperature (T) as:

λ = h/√2πmkT (1)

where h = Planck's constant = 6.626*10⁻³⁴ Js

m = mass of Rubidium = 85.47 amu * 1.66*10⁻²⁷ kg/ 1 amu = 1.419*10⁻²⁵ kg

k = Boltzmann constant = 1.38*10⁻²³ J.K⁻¹

T = temperature = 0.2 μK = 0.2 *10⁻⁶ K

Substituting these values in equation (1) we get:

λ = 6.626*10⁻³⁴ Js/√2π * 1.419*10⁻²⁵ kg * 1.38*10⁻²³ J.K⁻¹ * 0.2 *10⁻⁶ K

  = 4.224*10⁻⁷ m

Ans: The de Broglie wavelength is 4.224*10⁻⁷ m

amplitude

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