Helium Superfluid de Broglie Wavelength · CAP 2000 #14

Physics · quantum-mechanics

Problem

Helium 4^4He becomes a superfluid at temperatures T<2.18T < 2.18 K. A superfluid flows with no viscosity. This behaviour can only be explained using quantum physics and it can only happen if the de Broglie wavelength of a helium atom, of mass mm, is comparable to the inter-atomic spacing of the fluid. Which of the following could be an expression for λ\lambda, the de Broglie wavelength? (a) λ=h/3mkT\lambda = h/\sqrt{3mkT} (b) λ=3mkT/h\lambda = \sqrt{3mkT}/h (c) λ=h/3mkT\lambda = \sqrt{h/3mkT} (d) λ=3mkT/h\lambda = 3mkT/\sqrt{h}

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