Ratio of Excited Neon Atoms in 5s compared to 3p

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SUMMARY

The ratio of excited neon atoms in the 5s state compared to the 3p state at thermal equilibrium can be calculated using the Boltzmann distribution law. At a temperature of 300 K, with energy levels of E5s = 20.66 eV and E3p = 18.7 eV, the ratio N5s/N3p is calculated to be e-75.53, which is effectively zero. This indicates that virtually all neon atoms will occupy the 3p state, with almost none in the 5s state. The discussion emphasizes the importance of careful arithmetic and unit consistency when performing these calculations.

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jjson775
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Homework Statement
A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5s state to the number in a 3p state at 300 K? The energies of these states, relative to the ground state are E 5s = 20.66 eV and E 3p = 18.7 eV.
Relevant Equations
n5s/n3p = e ^-(E5 - E3)/kT
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"The ratio of the number of atoms in different energy states at thermal equilibrium can be calculated using the Boltzmann distribution law. The formula is:

N2/N1 = e-(E2-E1)/kT

where:
N2 and N1 are the number of atoms in the 5s and 3p states respectively,
E2 and E1 are the energies of the 5s and 3p states respectively,
k is the Boltzmann constant (8.6173 x 10-5 eV/K),
T is the temperature in Kelvin.

Substituting the given values:

N5s/N3p = e-(20.66 eV - 18.7 eV)/(8.6173 x 10-5 eV/K x 300 K)

N5s/N3p = e-1.96/0.02595

N5s/N3p = e-75.53

This ratio is extremely small, essentially zero for all practical purposes. Therefore, at 300 K, virtually all of the neon atoms will be in the 3p state, and almost none will be in the 5s state."

Found in less than 30 seconds using ck12.org/flexi (free access) :)
 
Thanks. I worked it in Joules and found my arithmetic mistake.
 
jjson775 said:
Homework Statement: A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5s state to the number in a 3p state at 300 K? The energies of these states, relative to the ground state are E 5s = 20.66 eV and E 3p = 18.7 eV.
Relevant Equations: n5s/n3p = e ^-(E5 - E3)/kT

View attachment 331584
What is your question? Are you asking us to check your work? Is that because it doesn’t give the answer you expect?

You need to be clear about what your are asking and it helps us to know why you are asking it.

Here are mistakes I spotted in your work. (There might be others.)

The expression ##e^{\frac {-(E_x – E_0)}{kT}}## makes no sense. Presumably you mean ##e^{\frac {-(E_5 – E_3)}{kT}}##.

##3.310 \times 10^{-18} - 2.996 \times 10^{-18}## is not ##0.335\times 10^{-18}##.

Your incorrect value of ##0.335\times 10^{-18}## divided by ##4.143 \times 10^{-21}## does not give ##8.06##.

The message is to check arithmetic carefully. Also, units are missing in various key places.

It's worth noting that you could simplify the arithmetic by looking-up and using the value of the Boltzmann constant (##k##) in units of eV/K.

Edit. Didn't see your Post #3 until after I'd posted.
 
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Correct. I was not used to using the Boltzmann constant in EV.
 

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