The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of 8 = kIn 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative?

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Percyaehyq · Accepted Answer

Given data:The energy of the first excited energy level of a hydrogen atom is U=10.2 eVThe entropy is S=kln4.The expression for the Helmholtz free energy of the first excited level is given by:  F  =  U  −  T  S    F  =  (  10.2  e  V  )  −  T  (  k  ln  ⁡  4  )At low temperature this is positive, so the atom would rather be in the ground state which has F = 0; at high temperature, however, the value of the F for the excited level becomes negative, so this level becomes preferred over the ground state, the transition temperature is where F = 0.Thus, the value of the temperature can be calculated as:  F  =  0    (  10.2  e  V  )  −  T  (  k  ln  ⁡  4  )  =  0    T  =            (      10.2      e      V      )              k      ln      ⁡      4          T  =            (      10.2      e      V      )              (      8.617      ×              10                  −          5                    e      V              /            K      )      ×      (      1.3863      )          T  =  8.539  ×      10    4  Thus, the value of the temperature is   T  =  8.539  ×      10    4

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-st

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