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

Click For Summary
The discussion focuses on calculating the ratio of excited neon atoms in the 5s state compared to the 3p state at thermal equilibrium using the Boltzmann distribution law. The formula involves the energies of the states and the Boltzmann constant, yielding a ratio that is extremely small, indicating that nearly all neon atoms are in the 3p state at 300 K. Participants point out arithmetic mistakes in the calculations and emphasize the importance of careful unit management and correct formula application. The conversation highlights the need for clarity in questions posed and encourages double-checking work for accuracy. Overall, the conclusion reinforces that at 300 K, virtually no neon atoms occupy the 5s state.
jjson775
Messages
112
Reaction score
26
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
1694020010691.jpeg
 
Physics news on Phys.org
"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.
 
Correct. I was not used to using the Boltzmann constant in EV.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
450
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
3
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K