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 3.371x10-19 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.

Answer 1

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

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

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

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

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

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