Why Can't I Get the Same Answer for the Boltzmann Equation Problem?

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SUMMARY

The forum discussion centers on discrepancies in calculating the Boltzmann equation for sodium atoms transitioning from the 3s to the 3p state. The key parameters include the constants Po = 2 and Pj = 6, with temperature values of 2500K and 2510K. The calculated energy difference, Ej, is derived from the average wavelength of emission lines at 589.3 nm, resulting in Nj/No = 1.72x10-4. The user is advised that their calculations appear correct, suggesting potential input errors in their calculator.

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  • Understanding of the Boltzmann equation and its application in statistical mechanics.
  • Familiarity with quantum states and energy transitions in atomic physics.
  • Proficiency in using scientific calculators or programming tools for exponential notation.
  • Knowledge of constants such as the Boltzmann constant (K = 1.38x10-23 J K-1) and Planck's constant (h = 6.6256x10-34 J s).
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  • Explore the use of scientific calculators or programming languages for handling exponential calculations.
  • Investigate the impact of temperature variations on atomic excitation and population ratios.
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Physicists, chemistry students, and anyone studying atomic behavior and statistical mechanics will benefit from this discussion, particularly those working with the Boltzmann equation and energy transitions in atoms.

atsum
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Why I can't calculate the same answer as the solution below? I use the value as what the below solution showed.

Q:
Where Nj is the number of atoms in excited state, No is the number of atoms in the
ground state, Pj and Po are constants determined by the number of states having equal
energy at each quantum level, Ej is the energy difference between excited and ground
states, K is the Boltzmann constant, and T is the absolute temperature. To understand
the application of this equation let us consider the situation of sodium atoms in the 3s
state (Po = 2) when excited to the 3p excited state (Pj = 6) at two different
temperatures 2500 and 2510K. Now let us apply the equation to calculate the relative
number of atoms in the ground and excited states:
Usually we use the average of the emission lines from the 3p to 3s where we have two
lines at 589.0 and 589.6 nm which is:

Solution:
λ = (589.0+589.6)/2 = 589.3 nm
Ej = hc/λ
Ej = (6.6256x10-34 J s x 2.998x1010 cm s-1)/{(589.3 nm x (cm/ 107 nm)}
Ej = 3.37x10-19 J
Nj/No = (6/2) exp {(-3.37x10-19 J/(1.38x10-23 J K-1 x 2500 K)}
Nj/No = 1.72x10-4
 
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atsum said:
Why I can't calculate the same answer as the solution below? I use the value as what the below solution showed.

Solution:
λ = (589.0+589.6)/2 = 589.3 nm
Ej = hc/λ
Ej = (6.6256x10-34 J s x 2.998x1010 cm s-1)/{(589.3 nm x (cm/ 107 nm)}
Ej = 3.37x10-19 J
Nj/No = (6/2) exp {(-3.37x10-19 J/(1.38x10-23 J K-1 x 2500 K)}
Nj/No = 1.72x10-4

Hard to tell where your problem is, calculations look correct and I got the same result from these numbers. Perhaps you are keying them wrong into calculator, or misunderstanding the notation. 2.998x1010 means 2.998x1010 and probably should be entered as 2.998e10 or something like that - details will depend on what (program, calculator model) you use.
 

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