Sodium emits many lines in the visible region of the e.m. spectrum. Tabulations for the
elements can be found in the Handbook of Chemistry and Physics [publ. Chemical Rubber
Co.] but it is not a primary resource which can be found in monographs published by NIST.
The dominant yellow emissions you are probably referring to are the doublet located at:
588.9950 nm and 589.5924 nm having a relative intensity of about 2/1.
Answered by: Vince Calder, Ph.D., Physical Chemist, retired
The wavelength of Sodium light actually consists of two wavelengths called the D lines.
1 = 588.995 nm
2 = 589.592 nm
These would easily be differentiated on a diffraction grating.
The angle of emergence, theta (), for a wavelength () is given by
d sin ()= n ()
So for each order n=1,2,... you will be able to see two lines very close to each other,
the higher wavelength being the more diffracted. The zero order will just be seen as one
In the experiment for each order to the left of the zero order fringe 1 would appear
on the right and 2 on the left while for each order to the right of the zero fringe
1 would appear on the left and 2 on the right.
Answered by: David Latchman, B.S., University of the West Indies, Trinidad
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