Street lights in many communities are sodium vapor lamps. These lamps visually h
ID: 496887 • Letter: S
Question
Street lights in many communities are sodium vapor lamps. These lamps visually have an orange yellow tint. The visible emission spectrum for sodium shows that two prominent bands are detected within the visible color These waves are both yellow and have wavelengths of about 590 nm. This explains why the light produced from sodium vapor lamps would appear yellow and not white. In comparison, mercury vapor lamps produce a blue colored light due to an intense light emission from the mercury atoms that has a wavelength of about 434 nm. a) Without doing any math, compare the wavelengths of the photon emitted from these two elements. Which element would you expect to produce higher energy photons in the visible part of spectrum? Discuss this with your group. b) Calculate the energy of the photon emitted by each element. Does your answer make sense given your initial expectation? c) In class you have learned that the light emissions from atoms are due to specific electron(s) movement between available energy levels. Would you expect that the size of the atom (the atom's radius) might also undergo changes as the atom releases photons with a specific color? Justify your answer below. Did your group agree or disagree initially?Explanation / Answer
a)
590 nm vs. 434 nm
comparison --> 590 > 434 longe wavelength, meaning, higher frequency for 434 nm, i.e. more energy
higher energy per photon --> that of 434 nm lamp, i.e. mercury lamp
b)
Apply math
E = hc/WL
where, h = planck constant 6.636*10^-34 and c = speed of light = 2.998*10^8 and WL = wavleneght in meters
so
E Hg = (6.636*10^-34)*(2.998*10^8) / (434*10^-9) = 4.58*10^-19 J/photon
E Na = (6.636*10^-34)*(2.998*10^8) / (590*10^-9) =3.371*10^-19 J/photon
this was as expected, the sodium lamp has less energy as mercury.
c)
Not likely, since this is simlpy excitation states of electrons in different shells/levels
the protons are not being involved