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ljucf
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Homework Statement
(a) What is the minimum kinetic energy in electron volts that an electron must have to be able to ionize a hydrogen atom (that is, remove the electron from being bound to the proton)? Answer: 13.6 eV
(b) If electrons of energy 12.8 eV are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?
Energy of highest-energy photon: 12.8 eV
Energy of next highest-energy photon: 12.089 eV
Energy of next highest-energy photon: 10.2 eV
Energy of next highest-energy photon: 2.6 eV
Energy of next highest-energy photon: 1.9 eV
Energy of lowest-energy photon: ?? eV
(c) If instead of electrons, photons of all energies between 0 and 12.8 eV are incident on a gas of hydrogen atoms in their ground state, what are the energies at which the photons are absorbed?
Energy of highest-energy dark line: 12.8 eV
Energy of next highest-energy dark line: 12.089 eV
Energy of lowest-energy dark line: 10.2 eV
Homework Equations
-13.6/n2
The Attempt at a Solution
I thought you would take 12.8-12.089 = .711 eV, however it states this is wrong. I have tried answers 0, .7, .711, and .6. Not sure why it's wrong.