Question

Ratio of population of states? Find the wavelength?

the ratio of population of two energy levels out of which upper one corresponds to a metastabe state is 1.059x10^-30. find the wavelength at 330K

Answer 1

I'm getting about `"631.7 nm"`.

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This can be gotten from statistical mechanics. The fraction of occupied states is:

`N_i/N = (g_ie^(-betaepsilon_i))/(sum_k g_ke^(-betaepsilon_k))`

where:

• `N_i` is the number of molecules populating state `i`, and `N` is the total number of molecules.
• `q_i = g_ie^(-betaepsilon_i)` is the partition function for state `i`.
• `g_i` is the degeneracy of state `i`.
• `beta = 1/(k_BT)` is a constant.
• `k_B = 1.38065 xx 10^(-23) "J/K"` is the Boltzmann constant.
• `T` is the temperature in `"K"`.
• `epsilon_i` is the energy of state `i`.

At a single temperature `T`, the ratio of the populations of states `i` and `j` is:

`(N_i"/"N)/(N_j"/"N) = N_i/N_j = ((g_ie^(-betaepsilon_i))/cancel(sum_k g_ke^(-betaepsilon_k)))/((g_je^(-betaepsilon_j))/cancel(sum_k g_ke^(-betaepsilon_k)))`

`= (g_ie^(-betaepsilon_i))/(g_je^(-betaepsilon_j)`

This ratio, `N_i/N_j`, was given as `bb(1.059 xx 10^(-30))`. This physically means that very little of the metastable state is actually occupied.

When the upper state `i` corresponds to a metastable state `i`, it indicates that `epsilon_i > epsilon_j`, since a metastable state is basically a temporarily stable state, and is thus more unstable than the ground state. Dividing through gives:

`= g_i/(g_j) e^(-betaepsilon_i - (-betaepsilon_j))`

Factor out the negative beta in the exponent to get:

`N_i/N_j = g_i/(g_j) e^(-beta[epsilon_i - epsilon_j])`

`= g_i/(g_j) e^(-betaDeltaepsilon)`

Assuming we are talking about singly-degenerate electronic states, we therefore assume that the degeneracies are both `1`, so that `g_i = g_j = 1`.

Then, the energy is analogous to the one you had been taught from general chemistry:

`Deltaepsilon = hnu = (hc)/(lambda)`

Therefore:

`N_i/N_j = e^(-Deltaepsilon"/"k_BT)`

`= e^(-hc"/"lambdak_BT)`

Now, we can solve for the wavelength.

`ln(N_i/N_j) = -(hc)/(lambdak_BT)`

`lambdaln(1.059 xx 10^(-30)) = -69.02lambda = -(hc)/(k_BT)`

`=> lambda = (hc)/(69.02k_BT)`

`= ((6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s"))/(69.02(1.38065 xx 10^(-23) "J/K")("330 K"))`

`= 6.317 xx 10^(-7)` `"m"`

This would be then, about `color(blue)("631.7 nm")`.