Capture by photon emission

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The interaction potential energy, U, between two atoms is shown in the figure at the right as a function of the distance separating them, r. The units chosen (eV - electron Volts - for energy, nanometers for distance) are common in describing the motion of atoms. The zero of potential energy is chosen so that when the atoms are far apart and at rest, their total mechanical energy is equal to zero.*

(a) If the atoms are far apart in a gas at room temperature, the kinetic energy of their relative motion will be about 15 meV (= millielectron Volts = 10-3 eV). If the two atoms approach each other with this energy, how close will they get to each other? Explain why you think so using a description of their motion and how you know this is what they do.

1155_Capturebyphotonemission.JPG

(b) If the two atoms come together in just the right way, they can lose energy by emitting a photon (a particle of light). Suppose that our atoms start with a total energy of 15 meV, approach each other and emit a photon that carries away 5 eV. Draw a line on the graph above that represents the total mechanical energy of the atoms.

(c) After the emission of the photon, are the two atoms bound (into a molecule) or free (they will fly apart)? When they move will there be a minimum or maximum distance they can reach? If so, explain why you think so and find those distances.

(d) If you answered "free" in part (c), what would have been the energy "lost" necessary to minimally bind them instead? If you answered "bound" in part (c), what would have been the energy "lost" that would have left them minimally free? Please explain as well as offer a numeric answer.

Reference no: EM13836649

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