The occupation probabilities of electrons in different states

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The occupation probability of electrons in different states is determined by the formula occup is proportional to [gi x exp(-Ei/kT)], where gi represents the number of states at energy Ei. A more comprehensive solution should include specific values for the number of states g_i for i = 1, 2, 3, along with a clear expression for the occupation probability that incorporates the proportionality constant. The term "temperature T6 = 1" refers to a normalized temperature scale, which needs to be converted to Kelvin for practical calculations. Understanding this relationship is crucial for obtaining accurate numerical results in thermodynamic contexts. The discussion emphasizes the importance of both the statistical mechanics framework and the correct temperature units in analyzing electron states.
Neo Tran
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Homework Statement
An atom with a single electron is in a heat bath at a temperature of T6 = 1. The atom is high Z, so the electron is bound at this temperature, and only three states have appreciable occupations. The ground state has spin 5/2. The first excited state, at 210 eV, has spin 3/2. The second excited state, at 380 eV, has spin 3/2. What are the occupation probabilities for these three states?
Relevant Equations
occup is proportional to [gi x exp(-Ei/kT)]
where gi is the numver of states at energy Ei
occup is proportional to [gi x exp(-Ei/kT)]
where gi is the numver of states at energy Ei
 
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How about a better attempt at a solution in which you write down (a) the number of states ##g_i## for ##i =1,2,3## and (b) an expression for "occup" that includes the proportionality constant?

Also, please explain what "temperature T6 = 1" means in terms of degrees K which is what counts when you need to find numerical answers.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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