Calculating Electron Distribution in Hydrogen Atoms

Click For Summary
To calculate the number of electrons in a specific energy level of hydrogen atoms at a given temperature, the Boltzmann factor and partition function must be utilized. At 3 Kelvin, the probability of electrons occupying the n=2 level is extremely low, primarily because the partition function is dominated by the ground state. This calculation is straightforward for gases, but for liquids, atomic interactions complicate the scenario significantly. The discussion reveals that the calculated probability for n=2 is exceedingly small, raising questions about the accuracy of the equations used. Overall, the calculations indicate that at very low temperatures, electrons are predominantly found in the ground state.
Rothiemurchus
Messages
203
Reaction score
1
How do I calculate the number of electrons with a given energy in
a sample of hydrogen atoms at a certain temperature?
For example how many electrons would be in the n=2 level
at 3 Kelvin? And does it matter whether or not hydrogen is gas or liquid or solid i.e is a different calculation needed for a gas in place of a liquid.
 
Physics news on Phys.org
This is a classical problem; you weigh each possible state by the Boltzmann factor e^(-E/kT), write the partition function and calculate the probability of a given energy level. At 3K, the probability of electrons being in n=2 is vanishingly small.

This only applies to a gas; in a liquid the interactions between atoms cannot be ignored and make for a messier problem.
 
When you say vanishingly small, have you got an approximate number
of decimal places for that small probability?
 
Easy to get. The partition function is dominated by the ground state so we can approximate:

Z = exp (-E1/kT).

P(n=2) = g * exp (-E2/kT) / Z, where g is the degeneracy of the 2nd level (ie how many distinct states there are with that energy level). Ignoring fine structure, g=4.

So P(n=2) = 4 exp (-(E2 - E1) / kT) = 4 e -39 455.3202 = 4 * 10 -17135

Hmm... that's even smaller than I expected. Did I do something wrong?
 
E1 = 10^-18 Joules.
P(n=2) = 4 exp (-(E2 - E1) / kT) = 4 exp [(4 x 10^-18 - 1 x 10^-18)/10^-23]

works out at about 10^10^-18

There must be something wrong with the equation.
 
Thread 'Unexpected irregular reflection signal from a high-finesse cavity'
I am observing an irregular, aperiodic noise pattern in the reflection signal of a high-finesse optical cavity (finesse ≈ 20,000). The cavity is normally operated using a standard Pound–Drever–Hall (PDH) locking configuration, where an EOM provides phase modulation. The signals shown in the attached figures were recorded with the modulation turned off. Under these conditions, when scanning the laser frequency across a cavity resonance, I expected to observe a simple reflection dip. Instead...

Similar threads

  • · Replies 83 ·
3
Replies
83
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • Sticky
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 0 ·
Replies
0
Views
487
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K