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a) This question requires use of the equation that relates the relative populations of states to the temperature and the energy difference between them;

\\[\\frac{N_u}{N_l}=e^{\\frac{-{\\Delta}E}{kT}}\\]

\\[{\\rm ln}(\\frac{N_u}{N_l})=\\frac{-{\\Delta}E}{kT}\\]

\\[kT{\\rm ln}(\\frac{N_u}{N_l})={-{\\Delta}E}\\]

\\[-(1.38~\\times~10^{-23}{\\rm~J~K^{-1}}\\times~\\var{temp}~{\\rm K}~\\times~{\\rm{ln}~}\\var{NuNl_rounded})=\\var{Energy_mantissa}\\times10^\\var{Energy_log}{\\rm J}\\]

b) To convert the energy of a photon into a frequency;

\\[E=h\\nu~\\]

\\[\\frac{E}{h}=\\nu=\\frac{\\var{Energy_mantissa}~\\times~10^\\var{Energy_log}~\\rm J}{6.626~\\times10^{-34}~\\rm~J~s}=\\var{Frequency_mantissa}\\times10^{\\var{Frequency_log}}~{\\rm Hz}\\]

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Calculate the energy difference, $\\Delta$E, between two states given that the relative populations of these states, $N_u/N_l$={NuNl_rounded} when the temperature, T, is {temp} K. The number of molecules in the upper state is $N_u$ and the number of molecules in the lower state is $N_l$;

${\\Delta}E$ = [[0]] $\\times$ 10^[[1]] J

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A transition occurs between the upper state and the lower state such that a photon is emitted. What is the frequency, $\\nu$, of the photon (in Hz)?

[[0]] $\\times$ 10^[[1]] Hz