Question

Laser Cooling of Atoms: In the simplest versions of laser cooling, a gas of atoms can be cooled from room temperature (?300 K) to about 1 × 10-4 K through the absorption of light. The absorbed light transfers momentum to the atoms, thereby slowing down those atoms that are moving in the direction of the incoming laser beam. The following questions refer to a gas of rubidium atoms.

(a) The average speed of atoms in the gas phase is directly related to the temperature of the gas through the relation (1/2)mv^2=(3/2)kT, where k is “Boltzmann’s constant”, with the numerical value 1.38 × 10-23 J/K. Calculate the average speed of a rubidium atom (m = 1.40 × 10-25 kg) at room temperature and at 1 × 10-4 K.

(b) Rubidium atoms absorb light at a wavelength of 780 nm (this is in the infrared). Individual photons of light of wavelength ? carry momentum according to the relation p=h/? where h is “Planck’s constant”, with the numerical value 6.63 × 10-34 J?s. What is the momentum of a single photon of 780 nm light?

(c) If a rubidium atom, approaching the laser beam, absorbs a single photon, by how much does the atom’s speed change?

(d) How many photons would such an atom need to absorb to be “cooled” from room temperature to 1 × 10-4 K?

Answer 1

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