Answer:
1) Probability that the molecule will be at the lowest energy state = 0.843
2) Probability that it will be the highest energy level = 0.000986
Explanation:
[tex]E = hc \bar{v}[/tex]
[tex]h = 6.67 * 10^{-34} \\c = 3 * 10^{8} \\hc = 2.001 * 10^{-25}[/tex]
The lowest energy is at [tex]\bar{v} = 0[/tex]
[tex]E_{1} = 2.001 * 10^{-25} * 0[/tex]
[tex]E_{1} = 0 J[/tex]
At [tex]\bar{v} = 500 * 10^{2} m^{-1}[/tex]
[tex]E_{2} = 2.001 * 10^{-25} * 500 * 10^{2}[/tex]
[tex]E_{2} = 1 * 10^{-20} J[/tex]
At [tex]\bar{v} = 1500 * 10^{2} m^{-1}[/tex]
[tex]E_{3} = 2.001 * 10^{-25} * 1500 * 10^{2}[/tex]
[tex]E_{3} = 3 * 10^{-20} J[/tex]
If the degeneracy of a system is g and the energy is E, the probability of the system is :
[tex]P(E) = \frac{ge^{\frac{-E}{kt} } }{\sum ge^{\frac{-E}{kt} } }[/tex]
Boltzmann constant, k = 1.380649×10−23 J/K.
Absolute temperature, T = 260 K
kT = 1.380649×10⁻²³ * 260
kT = 3.59 * 10⁻²¹
Probability that the molecule will be at the lowest energy state
Degeneracy g₁ = 0, g₂ = 3, g₃ = 5
[tex]P(E_{1} ) = \frac{1*e^{\frac{0}{kt} } }{ 1*e^{\frac{0}{kt} }+3*e^{\frac{-10^{-20} }{3.59*10^{-21} } }+5*e^{\frac{-3*10^{-20} }{3.59*10^{-21} } } }[/tex]
[tex]P(E_{1} ) = \frac{1}{1 + 0.185 + 0.00117}[/tex]
[tex]P(E_{1} ) =0.843[/tex]
2) Probability that it will be the highest energy level
[tex]P(E_{2} ) = \frac{5*e^{\frac{-3*10^{-20}}{3.59*10^{-21}} } }{ 1*e^{\frac{0}{kt} }+3*e^{\frac{-10^{-20} }{3.59*10^{-21} } }+5*e^{\frac{-3*10^{-20} }{3.59*10^{-21} } } }[/tex]
[tex]P(E_{2} ) = \frac{0.00117}{1+0.185+0.00117}[/tex]
[tex]P(E_{2} ) = 0.000986[/tex]
