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Suppose you have a two energy level system: $E_0=0$ and $E_1=\epsilon$ and $N$ molecules. We then have the number of molecules in the excited state to be:

$\displaystyle N_{\epsilon}(T)=\frac{N}{e^{\epsilon / k_bT}+1}$ Also: $\lim_{T\to \infty}N_\epsilon=N/2$.

Does this mean that we cannot add anymore energy to this system when $N_\epsilon=N/2$? Is there something which prohibits me from exciting another molecule to $\epsilon$?

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You can add energy to a state of N two-level molecules with infinite positive temperature. Then you will have states with negative temperature, which have higher energy. However, the total energy is still limited by $N\epsilon$.

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  • $\begingroup$ But I don't have lower energy states. Even if I did, these states release energy when they become occupied, not absorb. $\endgroup$ Feb 28, 2017 at 8:20
  • $\begingroup$ @RuskoRuskov: I am afraid I don't understand your comment or how it is relevant to my answer. You don't need lower energy states to get negative temperatures. $\endgroup$
    – akhmeteli
    Feb 28, 2017 at 13:16
  • $\begingroup$ "Then you will have states with negative energy." I do not have any $E_n<0$. A molecule either has energy $\epsilon$ or $0$. You mention negative temperatures. Negative temperatures are not associated with negative energies. Moreover in order to have negative temperature, $N_\epsilon>N/2$. Which is what I am asking: can you put more energy in the system even though $N_\epsilon=N/2$ $\endgroup$ Feb 28, 2017 at 18:30
  • $\begingroup$ @RuskoRuskov: You are right, that was a typo. I will edit it. $\endgroup$
    – akhmeteli
    Mar 1, 2017 at 1:54

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