In chemistry one occasionally uses a flame test to determine the presence of cer

Question

In chemistry one occasionally uses a flame test to determine the presence of certain chemicals. If we put a small spoon with sodium in into a Bunsen burner flame just above the tip (T 1500 K) it will glow with a characteristic yellow color. In this problem well work out about how many atoms of sodium are responsible for what we see. Well assume that as the sodium is heated, collisions with hot molecules drive the outermost electron of some of the sodium atoms up from its ground state (labeled 3s) to its first excited state (labeled 3p). The flame and the energy levels of the least bound electron in sodium are shown in the figure at the right. (Dont worry about the complexity of the diagram, we only need the two levels involved in the 3p-3s transition indicated by the yellow arrow.) The wavelength of the yellow light emitted by the 3p-3s transition in sodium is 590 nm. What is the energy of the emitted photon (in electron volts)? E_photon = eV Explain how you got your answer. At room temperature (T 300 K), what is the relative probability (compared to being in the ground state) that a sodium atom has its electron excited up to the 3p state? P_3p/P_3s = At the temperature of the flame (T 1500 K), what is the probability that a sodium atom has its electron excited up to the 3p state? P_3p/P_3s = Explain what physics you used to get your answers to parts B and C.

Explanation / Answer

For yellow light:

WL = 590 nm

so

E = hc/WL

h = plack constant = 6.636*10^-34

c = 3*10^8 speed of light

WL = 590*10^-9

then

E = (6.636*10^-34)(3*!0^8)/(590*10^-9) = 3.3742*10^-19 J/photon

change to eV

1 eV = 1.60218*10^-19 J

so

(3.3742*10^-19 / (1.60218*10^-19) = 2.1 eV

B)

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