Desirae Washington

2022-07-13

Can it be explained via blackbody radiation? Or is it purely to do with the material itself? For example, why does the line at 589.0 nm has twice the intensity of the line at 589.6 nm for sodium?

haingear8v

Sodium is not a blackbody. Rather the spectrum is determined by all the possible transitions between energy levels of it's electrons. Some of those levels are degenerate to varying degrees. If there are a lot of possible transitions with the same energy, the spectrum of the material will have a higher intensity for those energies, or wavelengths, respectively.

DIAMMIBENVERMk1

These two examples are of different origins.
The black body spectrum is the results of two factors: Bloltzmann distribuition ${e}^{-\beta \hslash \omega }$ and the available number of modes between $\nu$ and $\nu +d\nu$, $N\left(\nu \right)$. The intensity of spectrum of frequency $\nu$,
$I\left(\nu \right)=N\left(\nu \right)\sum _{j=1}^{\mathrm{\infty }}j{e}^{-j\beta h\nu }=\frac{N\left(\nu \right)}{{e}^{\beta h\nu }-1}\propto \frac{{\nu }^{3}}{{e}^{\beta h\nu }-1}$
The different intensity of Na (589 nm) and (589.6 nm) is due to the state multiplicity ${P}^{2/3}$ and ${P}^{1/2}$. The degeneracy is equal to $g\left(j\right)=2j+1$. For $j=\frac{3}{2}$ degeneracy $g\left(\frac{3}{2}\right)=4$, while $j=\frac{1}{2}$ degeneracy $g\left(\frac{1}{2}\right)=2$. The degeneracy ratio $g\left(\frac{3}{2}\right):g\left(\frac{1}{2}\right)=2:1$

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