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dedylja [7]
3 years ago
10

15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up

per state is one-third that of the lower state?
Chemistry
1 answer:
maksim [4K]3 years ago
3 0

Answer:

T = 525K    

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

\frac{N_{i}}{N} = e^{-\Delta E/kT}  (1)

<em>where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system </em>         

The energy between the two levels (ΔE) is:

\Delta E = hck    

<em>where h: is the Planck constant, c: is the speed of light and k: is the wavenumber</em>      

\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J  

Solving the equation (1) for T:

T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}  

<em>With Ni = N/3 and k = 1.38x10⁻²³ J/K, </em><em>the temperature of the two-level system is:</em><em> </em>

T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K                                  

I hope it helps you!

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