- #1
George Mahone
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[SOLVED] Probability Density
Consider a gas made of single hydrogen atoms (not diatomic hydrogen gas). The ground state energy of an electrons bound to a single hydrogen atom is -13.6 eV, and the energy of the first excited state is -10.2 eV. Ignoring the spin of the electron, the degeneracy of the ground state is 1, but the degeneracy of the first excited state if 4. (Note also that k = 8.617*10^-4 eV/K). What temperature would be required in order for the fraction of electrons in one specific orbital of the first excited states to be equal to the fraction of electrons in the ground state?I thought perhaps the electron probability densities would follow a canonical distribution, but that seems to be impossible.
Consider a gas made of single hydrogen atoms (not diatomic hydrogen gas). The ground state energy of an electrons bound to a single hydrogen atom is -13.6 eV, and the energy of the first excited state is -10.2 eV. Ignoring the spin of the electron, the degeneracy of the ground state is 1, but the degeneracy of the first excited state if 4. (Note also that k = 8.617*10^-4 eV/K). What temperature would be required in order for the fraction of electrons in one specific orbital of the first excited states to be equal to the fraction of electrons in the ground state?I thought perhaps the electron probability densities would follow a canonical distribution, but that seems to be impossible.