Consider a two-level system in which the excited state, separated from the ground state by energy ε, is doubly degenerate. The fraction of the molecules in the excited state, as T → ∞, is

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CSIR-UGC (NET) Chemical Science: Held on (15 Dec 2019)
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  1. \(\frac{1}{3}\)
  2. \(\frac{1}{2}\)
  3. \(\frac{2}{3}\)
  4. 1

Answer (Detailed Solution Below)

Option 3 : \(\frac{2}{3}\)
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Concept:

→ The partition function of a two-level system with doubly degenerate excited state is given by:

\(\frac{n_{i}}{N}=\frac{g_{i}e^{-E_{a}/KT}}{\Sigma g_{i}e^{-E_{a}/KT}}\)

\(\frac{n_{i}}{N}=\frac{g_{i}e^{-E_{a}/KT}}{g_{0}e^{-E_{a}/KT}+g_{1}e^{-E_{a}/KT}}\)

where k is the Boltzmann constant and T is the temperature.

The fraction of molecules in the excited state is given by:

\(f_{excited}=\frac{(2e^{(-\varepsilon /KT)})}{(1+2e)e^{(-\varepsilon /KT)}}\)

 

Explanation:

→ At T → ∞, we have \(\frac{1}{T}→ o\), which means that the denominator becomes much larger than the numerator. Therefore,

\(e^{\frac{-1}{T}}= e^{0} = 1\), due ton which exponential term will become 0.

\(\frac{n_{i}}{N}=\frac{2e^{-E/KT}}{1e^{0}+2e^{-Z/KT}}\)

When T → 

\(\frac{n_{i}}{N}=\frac{2}{1+2}=\frac{2}{3}\)

→ gexcited = 2. The ground state is non-degenerate, so gtotal = 2 (the degeneracy of the excited state) + 1 (the degeneracy of the ground state) = 3.

Conclusion:
Therefore, as T → ∞, the fraction of molecules in the excited state approaches 2/3.

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