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
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
1 Answer
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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
#(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,
When the upper state
#= 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
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