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Is Kinetic Energy of atom proportional to energy of its electons?


P_Rog

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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.

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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?

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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.

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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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