Question

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 upper state is one-third that of the lower state?

Answer 1

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)

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         

The energy between the two levels (ΔE) is:

$\Delta E = hck$    

where h: is the Planck constant, c: is the speed of light and k: is the wavenumber      

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

Solving the equation (1) for T:

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

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

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

I hope it helps you!