1 answer

Ratio of population of states? Find the wavelength?

Question:

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


Answers

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


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")#.

.

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