P_Rog Posted September 22, 2004 Posted September 22, 2004 Basically what the subject asks, with reference to Hydrogen gas. Is the kinetic energy of a hydrogen gas atom proportional to the energy of its electon or a photon emitted by an electron changing orbitals? I have a few notes on it, but i wanted a definent yes or no answer first.
fuhrerkeebs Posted September 22, 2004 Posted September 22, 2004 When you say the energy of it's electrons do you mean the total energy of the electrons or the just the kinetic energy?
fuhrerkeebs Posted September 22, 2004 Posted September 22, 2004 Then no...because an electron could have an outlier potential energy...
P_Rog Posted September 22, 2004 Author Posted September 22, 2004 ok, kinda a new question here, more of an interpretation: If i said that: (Kinetic Energy of a H atom at 300K) / ((orbital energy of H at n)(n^2)) = a constant, .0028499 for any n value and then (Kinetic Energy of a H atom at 200K) / ((orbital energy of H at n)(n^2)) = a constant, .001899948 for any n value and the on top of that: 200 * .0028499 = .56998 300 * .001899948 = .56998 Notice I'm multiplying Temp (200) with constant that was derived by using temp as 300 and vice versa for the other one. Does this mean anything? It would be better to elimate the certain temp and be able to use a variable of some sort to make it work for all cases. But does this mean anything? Basically i started out trying to relate the kinetic energy of a H atom with the energy of its orbitals, and since kinetic energy is based on temperature only (for gases) then does the temp affect the orbital energy?
P_Rog Posted September 22, 2004 Author Posted September 22, 2004 let me show some more of my work and what equations i used: for kinetic energy of an ideal gas = (3/2) * R * T where R is the gas constant (8.314) and T is temp in K take that number and divide by avogadro's number to get per atom energy of an orbital of H = R(sub h) * (h/n^2) where R(sub h) is the Rydberg constant (3.290 x 10^15), h is planck's constant (6.626 x 10^-34) and n is the energy state.
swansont Posted September 22, 2004 Posted September 22, 2004 The KE is not caused by the state of the electron, and is not, in general, caused by photon emission. But some fraction of the atoms will be in excited states based on the temperature of the system.
pulkit Posted September 22, 2004 Posted September 22, 2004 The first set of equations used come from classical mechnics, wherein you look at atoms as the smallest units of matter. The latter equations may be derived from quantum mechanics. Any mixing of these systems is generally not a good idea and would not give you any concrete results that you can be too happy with. The average kinetic energy of an atom has little to do with energy inside orbitals of electrons. You get a constt ratio because of the numerical form of the expression and constants involved.
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