Using a Boltzmann distribution, find the fraction of atoms in the excited state versus the ground state (i.e. the relative population) in a

Question

Using a Boltzmann distribution, find the fraction of atoms in the excited state versus the ground state (i.e. the relative population) in a plasma source and a flame source. Assume that the lowest energy of a sodium atom lies at 3.371×10-19 J above the ground state, the degeneracy of the excited state is 2, whereas that of the ground state is 1, and the temperature of the flame is 3000 K and 10,000 K for plasma.

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niczorrrr 4 years 2021-08-26T04:53:16+00:00 1 Answers 82 views 0

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  1. diemthu
    0
    2021-08-26T04:54:48+00:00
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    Answer:

    0.174 plasma

    $5.85 \times 10^{-4}$ flame

    Explanation:

    Given :

    Energy :

    $\Delta E=3.371 \times 10^{-19} $ J per atom

    $g^*=2$ (degenraci of excited state)

    g = 1  (degenraci of excited state)

    Boltzmann Distribution

    $\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$

    where,

    $N^*$ = atoms in excited state

    N = atoms in lower energy level

    k = $1.38 \times 10^{-23}$  J/K

    Therefore,

    Relative population in plasma

    T = 10,000 K

    $\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$

    $\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.37\times 10^{-19}}{1.38 \times 10^{-23} \times 10000}}$

    $\frac{N^*}{N}=2 \times e^{-2.44275}$

    $\frac{N^*}{N}=2 \times 0.8692$

    $\frac{N^*}{N}=0.1738$

    Relative population in flame    

    T = 3000

    $\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$

    $\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.371\times 10^{-19}}{1.38 \times 10^{-23} \times 3000}}$

    $\frac{N^*}{N}=2 \times e^{-8.1425}$

    $\frac{N^*}{N}=2 \times 2.9090 \times 10^{-4}$

    $\frac{N^*}{N}=5.85 \times 10^{-4}$

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