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Probability of molecular excited states

Page history last edited by Joe Redish 7 years, 2 months ago

7.4.P8

 

A diatomic molecule can vibrate, the two atoms oscillating back and forth. But by the rules of quantum physics, only very specific oscillations are permitted. For the molecule chosen, there are four possible vibrational states having energies E0E1E2, and E3. The energy separation between any two neighboring states is a fixed quantity ε. (Greek epsilon) That is En+1En = ε, for n = 0, 1, or 2. If this molecule is in thermal equilibrium in a fluid at temperature T, the probability of finding it in an state an energy ΔE above the ground state is 

 

where P0 is a normalization factor to guarantee that the probability of everything happening = 1. For simplicity, we write

1. What is the ratio of the probabilities of finding the molecule in its second excited state (E2)

compared to that of finding it in its ground state (E0)?

A. z     

B. z2    

C. z3    

D. z4    

E. Something else.

 

2. If the temperature increases and the system again reaches equilibrium,
the probability of finding the molecule in its second excited state

  1. Increases
  2. Decreases
  3. Remains the same
  4. Is impossible to determine from the information given.

 

3. If the temperature increases and the system again reaches equilibrium,
the probability of finding the molecule in its ground state

  1. Increases
  2. Decreases
  3. Remains the same
  4. Is impossible to determine from the information given.

 

4. If these four states are the only ones available to this degree of freedom,
then the normalizing factor for the probability, P0, must be

  1. 1
  2. 1/z
  3. 1/(1+z)
  4. 1/(1+z+z2)
  5. 1/(1+z+z2+z3)
  6. Something else
  7. Is impossible to determine from the information given.

 

 

 

Joe Redish 3/4/16 

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