Respuesta :
Answer:
a) The temperature is 48.41 K
b) P₁/P₀ = 1
c) PL/P₀ = 1
Explanation:
a) Given:
P₁/P₀ = 1/2
According the expression:
[tex]\frac{P_{1} }{P_{o} } =\frac{e^{-\beta E_{1} } }{e^{-\beta E_{o} } } \\\frac{P_{1} }{P_{o} }=\frac{e^{-\beta E_{1} } }{e^{-\beta *0 } }\\\frac{P_{1} }{P_{o} }=e^{-\beta E} \\\beta =\frac{1}{kt} \\\frac{P_{1} }{P_{o} }=e^-{\frac{\epsilon }{kt} }[/tex]
[tex]\epsilon =hf=6.626x10^{-34} *2.1x10^{12} =1.39x10^{-21} J[/tex]
k = Boltzmann constant = 1.38x10⁻²³
[tex]\frac{1}{2} =e^{\frac{-4.63x10^{-22} }{1.38x10^{-23} T} } \\\frac{1}{2}=e^{33.55T} \\ln(1/2)=-33.55/T\\-0.693=-33.55/T\\T=48.41K[/tex]
b) If T = 10%
[tex]T_{2} =0.1*48.41 =4.841K[/tex]
[tex]\frac{P_{1} }{P_{o} } =e^{\frac{-4.63x10^{-22} }{1.38x10^{-23} *4.841} }=1[/tex]
c) If
[tex]\frac{P_{L} }{P_{1} } =e^{-\beta (E_{2}-E_{1} } =ex^{-\frac{\epsilon }{kT} } \\E_{2}-E_{1} =2\epsilon -\epsilon = \epsilon \\Then\\\frac{P_{L} }{P_{1} } = 1 (same)[/tex]
The Boltzmann's equation allows to calculate the ratio of probabilities for the questions about particles in different state of energies are:
a) The state temperature is: T = 145.6K
b) For a temperature 0.1T the probability ratio is: P1 / Po =
c) The next ratio for the following two states is:
Boltzmann's equation establishes the probability of finding any particula within a specific energy state.
P = [tex]e^{- \frac{E_i}{kT}[/tex]
Where E is the energy of the state, k the Boltzmann constant and T the absolute temperature.
Part A
They indicate the frequency of oscillation f = 2.1 1012 Hz, they ask for what energy the ratio probability of finding the particle in the following state is:
[tex]\frac{P_1}{P_0} = \frac{1}{2}[/tex]
We substitute the Boltzmann's equation.
[tex]\frac{e^{- \frac{E_1}{kT} }}{e^{- \frac{e_0}{kT} }} = \frac{1}{2}[/tex]
[tex]e^{- \frac{(E_1-E_0) }{kT} } = \frac{1}{2}[/tex]
Planck's equation says that the energy of state is proportional to the frequency
E = h f
Where E is he enrgy, h the Planck's constant and f the frequency.
Let's substitute.
[tex]e^{- \frac{h f}{kT} } = \frac{1}{2}[/tex]
[tex]- \frac{hf}{kT} = ln \ 0.5[/tex]
T = [tex]- \frac{h f}{k \ ln \ 0.5}[/tex]
Let's calculate
T = [tex]- \frac{6.626 \ 10^{-34} \ 2.1 \ 10^{12} }{1.38 \ 10^{-23} \ ln \ 0.5}[/tex]
T = 1.45555 10²
T = 145.6K
Part b
Indicate that the temperature is 10% of the calculated temperature, which is the ratio of the probabilities.
The temperature is:
T = 0.1 145.6 = 14.56K
let's substitute to find the reason
[tex]\frac{P_1}{P_0} = e^{- \frac{hf}{kT } }[/tex]
[tex]\frac{P_1}{P_0} = e^{- \frac{6.626 \ 10^{-34} \ 2.1 \ 10^{12} }{1.38 \ 10^{-23} \ 14.56} }[/tex]
[tex]\frac{P_1}{P_0} = e^{- 6.9299}[/tex]
[tex]\frac{P_1}{P_0} = 9.8 \ 10^{-4}[/tex]
This indicates that almost all the particles are in the lowest state of energy.
Part c
For the same temperature, what is the probability ratio between state 2ε and state ε?
[tex]\frac{P_2}{P_1} = e^{- \frac{(E_2-E_1) }{k T} }[/tex]
The difference of energy between the states is:
E₂ - E₁ = ε
we substitute
[tex]\frac{P_2}{P_1} = e^{- \frac{\epsilon }{kT} }[/tex]
This value is the same as the calculation in part b, therefore the probability is:
[tex]\frac{P_2}{P_1}[/tex] = 9.8 10-4
In conclusion using the Boltzmann's equation we can calculate the probability ratio for the questions about the particles in different state of energies are:
a) The state temperature is: T = 145.6K
b) For a temperature of 0.1T the probability ratio is: [tex]\frac{P_1}{P_0} = 9.8 \ 10^{-3}[/tex]
c) The probability ratio for the following two states is: [tex]\frac{P_2}{P_1} = 9.8 \ 10^{-4}[/tex]
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