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
